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At leading order in the naive parton model, dimensionless deep-inelastic structure functions depend only on Bjorken x. Quantum chromodynamics evolution introduces scale dependence.

Work in natural units with $c=1$, neglect the electron mass, and put the initial proton at rest with mass $M$. Let $k$ and $k^{\prime }$ be the initial and final electron four-momenta, $q=k-k^{\prime }$, and $Q^2=-q^2>0$. Define the laboratory energy transfer ${\nu }_{\mathrm{lab}}=E-E^{\prime }=P\cdot q/M$. This ${\nu }_{\mathrm{lab}}$ has units of energy; the invariant $P\cdot q$ has units of energy squared. The Bjorken scaling variable and the final hadron invariant mass obeyThese follow by squaring $k-k^{\prime }$ and $P+q$, using $k^2=k^{\prime 2}=0$ and $P=(M,0)$. Solving the Bjorken x definition gives the outgoing energy and then the scattering angle:Alternatively, substituting $Q^2=4E{E}^{\prime }{\sin }^2(\theta /2)$ before solving gives $E^{\prime }=E/[1+2E{\sin }^2(\theta /2)/(Mx)]$. A chosen $x,E,Q^2$ must satisfy $E^{\prime }>0$ and $0\le {\sin }^2(\theta /2)\le 1$. In addition, $W$ must exceed the lightest allowed inelastic hadron threshold; positive $W^2$ alone is insufficient.

For an original numerical example, take $M=0.938\mathrm{GeV}$, $x=0.3$, $E=15\mathrm{GeV}$ and $Q^2=3{\mathrm{GeV}}^2$. Then ${\nu }_{\mathrm{lab}}\simeq 5.330\mathrm{GeV}$, $E^{\prime }\simeq 9.670\mathrm{GeV}$, $\theta \simeq 8.25^{\circ }$, and $W\simeq 2.807\mathrm{GeV}$. Varying $2\le Q^2/{\mathrm{GeV}}^2\le 5$ and $10\le E/\mathrm{GeV}\le 20$ at this same Bjorken x keeps every setting physical and gives $W^2>5.54{\mathrm{GeV}}^2$, above the proton-pion threshold and the stricter chosen continuum selection $W>2\mathrm{GeV}$. These are calculated settings, not historical measurements.

In the leading massless collinear parton approximation, $P^2=0$ and $q^2=-Q^2$ imply $(\xi P+q{)}^2=2\xi P\cdot q-Q^2$. With Bjorken x $x=Q^2/(2P\cdot q)$ and $P\cdot q>0$, the Dirac delta change-of-variable formula yields the displayed identity. The positive-energy branch of the final parton also gives $\delta (E_q+\xi E_P-E_k)/E_k=2\delta ((E_q+\xi E_P{)}^2-E_k^2)$ locally on physical support. Keeping target mass instead introduces ${\xi }^2{M}^2$ and changes the root and Jacobian.

The last panel of the preceding process figure shows deep inelastic scattering through a virtual photon. The blob represents the electromagnetic hadronic tensor response, with all final hadronic states included. With charges factored into $e$ and the hadronic current as in the question, a suitable scattering amplitude, up to an irrelevant overall phase, isThe electron fermion spin sum, before averaging the two initial spin states, gives the following normalization of the leptonic tensor:The initial electron spin average is a separate factor $1/2$. Equivalently one can absorb it into a tensor $L/2$, but the prefactor in the requested formula uses the unaveraged trace $L$. The massless Dirac equations imply $q^{\mu }{L}_{\mu \nu }=q^{\nu }{L}_{\mu \nu }=0$, the relevant Ward identity.

In the target rest frame the incident flux is $4p\cdot P=4ME$. This remains valid for a massive target because the electron is massless: the general flux $4\sqrt{(p\cdot P{)}^2-p^2{P}^2}$ reduces to that expression. The outgoing electron contributes $d^3{p}^{\prime }/((2\pi {)}^{3}2E^{\prime })$, and the inclusive final-state sum is $4\pi W_H^{\mu \nu }$. Including the electron spin average therefore givesThe implicit hadron spin average in $W_H$ must not be inserted a second time.

For an unpolarized target, Lorentz invariance permits tensor structures formed from $P$, $q$ and the metric. Hermiticity relates the two index orders. The electromagnetic response is parity even, excluding the pseudotensor $i{ϵ}^{\mu \nu \rho \sigma }{P}_{\rho }{q}_{\sigma }$. A possible ordinary antisymmetric $P^{\mu }{q}^{\nu }-q^{\mu }{P}^{\nu }$ term is excluded by current conservation. The symmetric part initially has four structures, $g^{\mu \nu }$, $P^{\mu }{P}^{\nu }$, $q^{\mu }{q}^{\nu }$ and $P^{\mu }{q}^{\nu }+q^{\mu }{P}^{\nu }$. Imposing both current-conservation conditions leaves the transverse decomposition of an unpolarized hadronic tensor:For example, contracting $q$ into the first structure gives $(1+A)q^{\mu }$ in the printed parametrization, and contracting into either parenthesis gives $(1+B)\nu$ or $(1+C)\nu$. Since the two independent deep-inelastic structure functions are not generally linked by symmetry, transversality requiresThere are only two invariant variables once $P^2=M^2$ is fixed; they may be chosen as $Q^2$ and Bjorken x $x=Q^2/(2\nu )$.

Now make the leading massless collinear parton approximation. In the hard parton calculation, use a lightlike collinear reference for $P$, neglecting target-mass effects of relative order $M^2/Q^2$ as well as quark masses. Literally combining a massive $P$ with an exactly massless constituent momentum $\xi P$ would be inconsistent. The exact rest-frame flux derived above does not require this approximation; it is the following partonic response that does.

Let $r=\xi P$ and $k=r+q$. Strip the quark charge out of the current, since its square is already the factor $Q_f^2$ in the parton convolution. Averaging the initial quark spin and summing the final spin gives the spin-averaged electromagnetic parton tensorThe spatial momentum delta fixes $\mathbf{k}=\mathbf{q}+\xi \mathbf{P}$. The remaining one-particle Lorentz-invariant phase space, together with $1/(4\pi \xi )$ in the definition, suppliesSince $r^2=k^2=0$ and $r\cdot k=\xi P\cdot q$, the full parton response isTerms with a free $q$ index vanish when contracted with the conserved leptonic tensor. Thus the part of $\tilde{W}$ contributing to the cross-section iswhere $\doteq$ denotes equality in that contraction, rather than an equality of the full tensors. The omitted terms are necessary for the full hadronic tensor to be transverse; the wording about contributing terms in the PDF is important here.

To prove the delta identity with the correct Jacobian, set $\omega =E_q+\xi E_P$. On physical support $E_k>0$ and $\omega >0$. Factoring $E_k^2-{\omega }^2=(E_k-\omega )(E_k+\omega )$ gives the positive-energy on-shell delta function identityThis is restricted to the positive-energy branch; the other root must not be included. Using the spatial momentum constraint and $P^2=0$ in the leading hard approximation,The Dirac delta function change of variable therefore gives the parton on-shell delta identityHere $\nu >0$. This is the actual printed identity: the converted TeX's final $\delta (x)/E_k$ is a transcription error. If target mass is formally retained in the delta argument, it becomes $Q^2-2\xi \nu -{\xi }^2{M}^2$, so the stated simple root and Jacobian no longer apply.

Let $D(x)={\sum }_f{Q}_f^2(q_f(x)+{\overline{q}}_f(x))$. The parton integral collapses at $\xi =x$, givingThe transverse decomposition of an unpolarized hadronic tensor, in the same contraction, is $-g^{\mu \nu }{F}_1+P^{\mu }{P}^{\nu }{F}_2/\nu$. Comparing the two independent structures yieldsThis derives the Callan-Gross relation from the spin-one-half trace and on-shell constraint. Antiquarks give the same symmetric response and therefore the sum $q_f+{\overline{q}}_f$. The result applies at leading order in the massless parton model; target masses, intrinsic transverse momentum and Quantum chromodynamics radiative effects can generate a nonzero longitudinal deep-inelastic structure function and modify the relation.

Use an incoming spacelike momentum transfer $q$, so the physical absorption process has $p_X=P+q$, $Q^2=-q^2>0$, and $P\cdot q>0$. There is a sign inconsistency in the original PDF: its delta function has $P-p_X-q$, while its positive Bjorken x definition and target formulas use incoming $q$. With that literal delta function and $x>0$, the final-state invariant would bewhich becomes negative in the stated high-energy limit and cannot represent a physical hadronic final state. We therefore use ${\delta }^4(P+q-p_X)$, preserving the printed $x=Q^2/(2P\cdot q)$ and structure-function conventions. Equivalently, keeping an outgoing momentum variable requires reversing the signs in those definitions as well.

Assume an unpolarized target, incoherent scattering from effectively free spin-one-half partons, and the massless collinear parton approximation. In the Bjorken scaling limit, take $Q^2,P\cdot q\to \infty$ at fixed $0<x<1$, neglecting target and quark masses and intrinsic transverse momentum relative to $Q$. Write the incoming struck-quark momentum as $k=\xi P$ in the leading lightlike-reference approximation. Let $f_f(\xi )$ and ${\bar{f}}_f(\xi )$ be quark and antiquark parton distribution functions, counting all colours. For the naive scaling result also neglect radiative scaling violations; in full QCD these distributions depend logarithmically on the hard scale.

The outgoing struck quark has momentum $k^{\prime }=k+q$. Its on-shell condition givesAverage its initial spin and sum its final spin. The spin-averaged electromagnetic parton tensor has numeratorThe supplied three-gamma identity gives this directly: contract its middle index with $(k+q{)}_{\lambda }$ and trace with $/k$; the ${\gamma }_5$ trace vanishes. Thus no antisymmetric term survives the unpolarized spin average.

Define transverse tensorsOn the cut, $2k\cdot q=Q^2$, so $\tilde{k}=k+q/2=\xi \tilde{P}$. Expanding the two transverse terms givesThis also explicitly verifies electromagnetic current conservation, $q_{\mu }{S}^{\mu \nu }=q_{\nu }{S}^{\mu \nu }=0$.

The one-particle final-state Lorentz-invariant phase-space measure obeysIncluding the current's charge factor and the given $1/(4\pi )$ normalization, a single quark therefore contributesAn antiquark has the same unpolarized tensor and the same squared charge $Q_f^2$.

The parton convolution normalization for a hadronic tensor isThe $1/\xi$ is important: the distributions count parton numbers, whereas covariantly normalized one-particle responses carry the state normalization $2k^0$. The incoherent sum of normalized responses is $W/(2P^0)={\sum }_f\int d\xi (f_f+{\bar{f}}_f)w_f/(2k^0)$; in the collinear frame $k^0=\xi P^0$, giving the factor above. There is no extra colour multiplicity because it is already included in the distributions.

On the physical cut,Comparing the two transverse coefficients in the hadronic tensor gives, with $n_f=f_f+{\bar{f}}_f$,Therefore the requested deep-inelastic structure functions areThe notation $f_f$ here is the question's $f$ with its flavour index made explicit. The calculation proves the Callan-Gross relation, $F_2=2x{F}_1$, rather than merely citing it. Finite masses and radiative corrections modify this leading approximation; at leading QCD order the same formulas use scale-dependent parton distribution functions.

The quark number sum rule follows because a distribution is a number density. Integrating $f_f$ counts quarks and integrating ${\bar{f}}_f$ counts antiquarks; their difference is the net conserved flavour charge:Every sea quark-antiquark pair contributes zero to this difference. More formally the flavour-vector charge counts quarks minus antiquarks and has a fixed eigenvalue on the hadron state. Thus $N_f$ is the valence or net quark number, not the total number including the sea.

Use both final assumptions printed in the question: only up/down flavours and no antiquarks. The second does not follow just from restricting the flavour list. Then $\int f_u=N_u$, $\int f_d=N_d$, with squared charges $Q_u^2=4/9$ and $Q_d^2=1/9$. The proton has $(N_u,N_d)=(2,1)$ and the neutron $(1,2)$. Hence the valence-only proton and neutron structure-function moments areThe integral limits assume the scaling approximation is valid for the indicated weighted moments, for example by dominated convergence of the weighted integrands. The same net-quark sum rule alone would not fix these individual moments in the presence of antiquarks, since $F_2$ involves the sum of quark and antiquark distributions.

For an unpolarized electromagnetic target, Lorentz covariance and parity symmetry in quantum field theory permit a symmetric hadronic tensor built from $g^{\mu \nu }$, $P^{\mu }$ and $q^{\mu }$, where $P=P_H$. Its surviving general form isThe totally antisymmetric tensor structure ${ϵ}^{\mu \nu \rho \sigma }{P}_{\rho }{q}_{\sigma }$ is excluded by electromagnetic parity symmetry in quantum field theory for this unpolarized target. The other possible antisymmetric structure, $P^{\mu }{q}^{\nu }-q^{\mu }{P}^{\nu }$, cannot be transverse for generic scattering kinematics and hence is excluded by current conservation. Current conservation gives the Ward identities $q_{\mu }{W}^{\mu \nu }=q_{\nu }{W}^{\mu \nu }=0$, imposingTwo scalar functions remain. Taking ${\tilde{P}}^{\mu }=P^{\mu }-\nu q^{\mu }/q^2$ gives the transverse basisThese are the deep-inelastic structure functions; $x$ is Bjorken x. At fixed target mass, the two independent Lorentz scalar invariants can equivalently be chosen as $x,q^2$. In this paper $\nu =P\cdot q$ has dimensions of mass squared, rather than the alternative convention $\nu =P\cdot q/M$ used for energy transfer.

An ideal Bjorken scaling model has $F_2(x,Q^2)=f(x)$: varying $Q^2$ at fixed Bjorken x leaves this dimensionless deep-inelastic structure function unchanged. The comparison requires extracting a deep-inelastic structure function, rather than plotting raw event counts. Quantum chromodynamics predicts departures from exact Bjorken scaling; finite hadron masses also limit the idealization.

For an original shape comparison, choose $f(x)=x(1-x{)}^3$ and define $S(Q^2)=F_2(0.3,Q^2)/F_2(0.3,Q_0^2)$ with $Q_0^2=2{\mathrm{GeV}}^2$. The ideal model has $S=1$. To display a contrasting size-dependent suppression, define an illustrative elastic-form-factor shape $G(Q^2)=(1+Q^2/{\Lambda }^2{)}^{-2}$ with the freely chosen ${\Lambda }^2=1{\mathrm{GeV}}^2$. Its normalized squared magnitude isThus $A(5{\mathrm{GeV}}^2)=1/16$, whereas $S(5{\mathrm{GeV}}^2)=1$. The elastic-form-factor shape is a comparison of suppression mechanisms, not an elastic differential scattering cross-section assigned to $x=0.3$: actual elastic scattering from a proton has $W=M$ and $x=1$. Neither curve is measured data or a fit to the SLAC-MIT experiment.

The experimental interpretation required more than a flat curve. The angular dependence and small longitudinal response supported spin one-half partons, expressed at leading order by the Callan-Gross relation $F_2=2x{F}_1$. Identification with fractionally charged quarks also drew on later electron, muon and neutrino comparisons. A scaling plot alone does not determine the individual electric charges of constituents.^[ref]

The leading parton model formula already shows the ambiguity: $F_2=x{\sum }_f{Q}_f^2(q_f+{\bar{q}}_f)$ combines squared electric charges with unknown parton distribution functions. One measured function cannot independently fix all these quantities. Confinement also explains why an inclusive final hadron state need not contain an isolated outgoing quark.