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Here $Q^2=-q^2>0$. In the leading massless collinear parton approximation, this kinematic variable equals the momentum fraction of the struck parton.

For massless spin-one-half partons at leading order, the unpolarized electromagnetic deep-inelastic structure functions obey this relation. Radiative Quantum chromodynamics corrections and target-mass effects can modify it.

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.

On the physical positive-energy branch, put $\omega =q^0+\xi E_P$. The massless outgoing parton obeysSince $E_{k^{\prime }}=\omega >0$ on the support and $\nu >0$, the positive-energy on-shell delta function identity givesInsert this into the parton model sum, and writeThe parton distribution functions $q_f,{\bar{q}}_f$ are number densities in momentum fraction. The contributing tensor becomesComparing with part (b), under the same leptonic tensor contraction, givesThis is the Callan-Gross relation for massless partons of spin angular momentum $1/2$ at leading order. The longitudinal deep-inelastic structure function is $F_L=F_2-2x{F}_1=0$ in this approximation. Target-mass effects and radiative Quantum chromodynamics corrections can change the relation. The leading parton model also gives Bjorken scaling: at this level the deep-inelastic structure functions depend on $x$ alone.

Use the massless collinear parton approximation in a high-energy frame: $P^2=0$, $k=\xi P$, and $k^{\prime }=\xi P+q$ with $k^{\prime 2}=0$. This neglects target-mass corrections to the parton model; it does not literally set a stationary massive target to a massless particle in the earlier flux formula.

For a quark of dimensionless charge $Q_f$, the electromagnetic vector current matrix element is $Q_f\bar{u}(k^{\prime }){\gamma }^{\mu }u(k)$. The spin average and gamma-matrix trace giveIntegrating the three-momentum Dirac delta function in the parton hadronic tensor leavesSince $k\cdot k^{\prime }=\xi \nu$, this isFor the massless Electron momenta, $p^2=p^{\prime 2}=0$ and $q=p-p^{\prime }$. Substitution into the leptonic tensor givesand likewise $q^{\nu }{L}_{\mu \nu }=0$. These Ward identities eliminate every term with an exposed $q$ index in the contraction. ThereforeHere $\doteq$ means equality after contraction with the leptonic tensor. The shortened tensor is not itself conserved; the omitted terms restore current conservation in the full hadronic tensor.