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For the current with $(1-{\gamma }^5)$ rather than $P_L$, the spin-summed symmetric leptonic tensor is $8(q_{1\mu }{q}_{2\nu }+q_{1\nu }{q}_{2\mu }-g_{\mu \nu }{q}_1\cdot q_2)$. Integrating the standard two-body Lorentz-invariant phase space of two massless particles givesIts antisymmetric Levi-Civita symbol term integrates to zero. The normalization changes by a factor of four if the current uses $P_L$ instead. The result separates the leptonic integration from hadronic pseudoscalar-to-pseudoscalar form factors in a three-body decay rate.

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.

Let $s=q^2>0$, neglect $m_e$, and average over the two spin states of each incoming particle. Only the single-photon channel is being considered. The spectral-density symbol ${\rho }_h$ is not defined in the question, so specify its normalization before calculating. A useful photon-field convention isHere $d{\Pi }_X$ is the product of final-particle measures $d^{3}k/[(2\pi {)}^{3}2E]$, including the appropriate sums and symmetry factors. Lorentz invariance and current conservation give the transverse tensor form. In this convention ${\rho }_h$ is the hadronic photon spectral density, with mass dimension $-2$, rather than a density of a stationary time series.

The electron annihilation amplitude can be written $e\bar{v}(p_{+}){\gamma }_{\mu }u(p_{-})⟨X|A^{\mu }|0⟩$, up to an overall phase. The leptonic tensor after the initial spin average isIt obeys $q^{\mu }{L}_{\mu \nu }=0$ and $g^{\mu \nu }{L}_{\mu \nu }=-s$. In the center-of-momentum frame, the flux denominator in the hint is $2s$, since the relative speed is 2 and each beam energy is $\sqrt{s}/2$. Contracting the tensors gives the inclusive relativistic cross-sectionHere $\alpha$ is the fine-structure constant. The $1/4$ in the initial spin average is essential because the hint's particles were spinless.

For the other common convention, define the hadronic electromagnetic current with the coupling omitted, $J_h^{\mu }={\sum }_f{Q}_f{\bar{q}}_f{\gamma }^{\mu }{q}_f$, and write its inclusive tensor asEquivalently, for $i\int d^{4}x{e}^{iqx}⟨0|T{J}_h^{\mu }(x)J_h^{\nu }(0)|0⟩=(q^{\mu }{q}^{\nu }-q^2{g}^{\mu \nu }){\Pi }_J(q^2)$, one has ${\rho }_J=\mathrm{Im}{\Pi }_J/\pi$. At leading order in the electromagnetic interaction, $⟨X|A^{\mu }|0⟩$ is proportional to $e⟨X|J_h^{\mu }|0⟩/s$. ThusIf the symbol ${\rho }_h$ is instead used for this dimensionless hadronic electromagnetic-current spectral density, the last formula applies with ${\rho }_J$ renamed ${\rho }_h$. The normalization must not be silently switched between these formulas.

At $\sqrt{s}$ well above the strong-coupling scale of Quantum chromodynamics, asymptotic freedom makes production over distances of order $1/\sqrt{s}$ perturbative. The electromagnetic current initially creates a quark-antiquark pair. Subsequent strong interactions produce hadrons, but an inclusive sum over all hadronic final states is much less sensitive to this rearrangement than an exclusive channel. This is the regime in which quark-hadron duality motivates a leading parton model calculation, with radiative and power-suppressed corrections. It is not a pointwise theorem at individual resonances or near thresholds; the comparison is most reliable for sufficiently inclusive or suitably averaged high-energy observables. This argument remains restricted to photon exchange, even where additional electroweak channels could also contribute.

For one active massless quark flavor with charge $Q_{f}e$, the tree-level scattering amplitude isThe two gamma-matrix traces, the initial spin average, and the final color charge sum yieldwhere $t=-s(1-\cos \theta )/2$ and $u=-s(1+\cos \theta )/2$. The massless two-body Lorentz-invariant phase space then givesSumming the distinct final flavors, rather than interfering amplitudes for them, givesAccording to the permitted approximation in the paper, active flavors satisfy $m_f^2<s$ and are treated as massless; the others are omitted. This is the stipulated step approximation, not the exact pair-production threshold $s\ge 4m_f^2$.

The hadronic R ratio is $R=N_c\sum Q_f^2$, relative to the massless muon-pair relativistic cross-section $4\pi {\alpha }^2/(3s)$. If $N_u$ and $N_d$ active flavors have charges $2/3$ and $-1/3$, respectively,For $N_c=3$, the usual sets of three, four, five and six active flavors give $R=2,10/3,11/3,5$. In the two spectral conventions the same leading calculation gives

The strong-interaction matrix element between two spin-zero pseudoscalar mesons has only $p^{\mu }$ and $k^{\mu }$ available. The product of the two intrinsic parities is positive. An axial current would require a pseudovector constructed from these momenta, but an expression involving the Levi-Civita symbol needs three independent four-vectors and therefore vanishes. This is a consequence of parity conservation in the hadronic matrix element, not of parity conservation in the weak interaction. The vector current can have the two independent structures $p+k$ and $p-k$. Its coefficients are Lorentz scalars; with $p^2=m_K^2$ and $k^2=m_{\pi }^2$ fixed, their only varying invariant is $s=q^2$. HenceThese are the pseudoscalar-to-pseudoscalar form factors. With relativistically normalized states they are dimensionless. The vanishing axial matrix element and this decomposition explain the two equalities separately.

Write $q_1$ and $q_2$ for the outgoing electron and antineutrino momenta. From the Fermi interaction, an invariant scattering amplitude, up to an irrelevant overall sign or phase, isThe CKM matrix element $V_{us}$ multiplies the quark current in the convention of this interaction. Let ${\ell }_{\mu }=\bar{u}(q_1){\gamma }_{\mu }(1-{\gamma }^5)v(q_2)$. For massless leptons, the massless Dirac equation and chirality matrix anticommutation giveIn the second term move $/q_2$ through the chiral projector before applying $/q_{2}v=0$. Since $p+k=2p-q$, this transverse massless leptonic current givesThe disappearance of $f_{-}$ uses the massless approximation; for a massive charged lepton its contraction is proportional to the lepton mass.

Use the fermion spin sum and the supplied gamma matrix trace identities. The symmetric part of the leptonic tensor isThe Levi-Civita symbol term is antisymmetric and drops out when contracted with $p^{\mu }{p}^{\nu }$. ThusThere is no initial-spin average because the kaon is spinless. For the integrated massless leptonic tensor, keep every factor of $2\pi$ explicit and define the unnormalized two-lepton Lorentz-invariant phase spaceThe leptons are massless, so $q_i^0=|{\mathbf{q}}_i|$. Its trace is $g^{\mu \nu }{I}_{\mu \nu }=\pi s$. ThereforeThe three Lorentz-invariant phase-space measures and their momentum delta function contribute $1/[8(2\pi {)}^5]$, in addition to $1/(2m_K)$ in the decay rate. Combining them with the spin sum givesThis massless semileptonic pseudoscalar decay rate uses a two-lepton integral over future-timelike $q$ and the pion integral is restricted to the physically allowed region. The null endpoint follows by continuity. The coefficient has mass dimension $-5$, so the complete expression has mass dimension one, as a decay rate must in natural units.

In the kaon centre-of-momentum frame, put $E=k^0$ and $\kappa =|\mathbf{k}|$. ThenThe dimensionally consistent Källén function isThe pion-only mass term must have fourth power: the second power printed in the PDF is dimensionally inconsistent. This repair also follows directly from squaring $E=(m_K^2+m_{\pi }^2-s)/(2m_K)$. Angular integration and the change of variable givewhere the negative sign reverses the endpoints. Combining this with the bracket $\lambda /4$ yieldsThus $\boxed{B=G_F^2|V_{us}{|}^2/(192{\pi }^3{m}_K^3),a=0,b=(m_K-m_{\pi }{)}^2}$. The lower limit is the minimum invariant mass of two massless leptons; at the upper limit the pion is at rest. The coefficient has mass dimension $-7$, while $ds{\lambda }^{3/2}$ has dimension eight. The Källén function also shows why the differential decay rate vanishes at zero pion momentum.

The deep inelastic scattering process contains a virtual photon exchanged between the electron and the hadron:

Use an electromagnetic vector current $J^{\mu }$ containing the dimensionless quark charges, with the coupling $e$ factored out. The scattering amplitude, up to an irrelevant phase, isTo match the printed prefactor, define the leptonic tensor with a spin sum over both electron spins and keep the initial spin average $1/2$ outside it. The gamma-matrix trace givesFor a stationary target and massless electron, the invariant flux factor is $4ME$. The inclusive final-state Lorentz-invariant phase-space measure and target spin average are contained in $4\pi W_H^{\mu \nu }$. Thus the differential scattering cross-section isHere $q^4$ means $(q^2{)}^2$. If the initial spin average is instead built into the leptonic tensor, its normalization is $2(\cdots )$ and the displayed cross-section prefactor must be doubled. The two conventions give the same observable.

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.

For electromagnetic deep inelastic scattering with an unpolarized target, Lorentz invariance, parity and current conservation leave two independent deep-inelastic structure functions. With $T^{\mu \nu }=g^{\mu \nu }-q^{\mu }{q}^{\nu }/q^2$ and $\tilde{P}=P-(P\cdot q/q^2)q$, both tensor structures are transverse to $q$. Terms with a free $q$ index disappear on contraction with a conserved leptonic tensor, although such terms can still be necessary to make the full hadronic tensor transverse.