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The Källén function controls two-body relativistic kinematics. In the rest frame of total invariant mass $\sqrt{s}$, the magnitude of each final momentum is $\sqrt{\kappa (s,m_1^2,m_2^2)}/(2\sqrt{s})$. It also determines the invariant flux factor in a two-particle collision.

Use metric $(+---)$, ${ϵ}^{0123}=+1$ and relativistically normalized external spinors. The Fermi interaction givesup to an irrelevant phase. The fermion spin sums replace $u\bar{u}$ and $v\bar{v}$ by the slashed massless momenta. Write $C=P^2+Q^2$. The gamma matrix trace identities yieldThe symmetric-antisymmetric mixed contractions vanish. The symmetric contraction is $2[(a\cdot c)(b\cdot d)+(a\cdot d)(b\cdot c)]$, while the two epsilon tensors contract to $-2[(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)]$. Including $G_F^2/2$ and the $1/4$ initial-spin average therefore givesThe singlet color charge contractions ${\delta }_{ij}{\delta }_{kl}$ have net factor one after summing final colours and averaging initial color charge states. Thus this is also the colour-averaged partonic result; no extra factor three is required.

In the centre-of-momentum frame, let $c_{\theta }=\cos \theta$. The Mandelstam variables satisfy $t=-s(1-c_{\theta })/2$ and $u=-s(1+c_{\theta })/2$, giving $(p_2\cdot k_1)(p_1\cdot k_2)=s^2(1+c_{\theta }{)}^2/16$ and $(p_2\cdot k_2)(p_1\cdot k_1)=s^2(1-c_{\theta }{)}^2/16$. HenceThe massless invariant flux factor is $2s$ and $d{\Phi }_2=d\Omega /(32{\pi }^2)$, so the weak charged-current quark scattering cross-section isFor a purely left-handed weak charged current, $P=1$, $Q=-1$, this reduces to $d\sigma /d\Omega =G_F^{2}s(1+\cos \theta {)}^2/(32{\pi }^2)$, fixing the sign of the forward term. Purely right-handed weak charged currents at both vertices have the same unpolarized angular law. The factorization convention for $F,H_1,H_2$ could be rescaled by a common numerical factor; the displayed product fixes the normalization.

Use the Lorentz-invariant phase-space measure from the original PDF, including the $1/(2E_{p^{\prime }})$ and $1/(2E_{q^{\prime }})$ factors missing from the local TeX. In the center of mass frame put $E=\sqrt{s}/2$ and $k=\sqrt{s-4m^2}/2$, assuming $s>4m^2$. The spatial Dirac delta distribution sets ${\mathbf{q}}^{\prime }=-{\mathbf{p}}^{\prime }$, leavingSince $d(2E_{k^{\prime }})/d{k}^{\prime }=2k^{\prime }/E_{k^{\prime }}$, the radial integral givesThe invariant flux factor is $2\sqrt{\kappa (s,m^2,m^2)}=2\sqrt{s(s-4m^2)}$, using the Källén function. Dividing the phase space by this flux yields $d\sigma /d\Omega =|\mathcal{M}{|}^2/(64{\pi }^{2}s)$, with the final-state labels retained as in the printed formula.

Let $x=\cos \theta$. Of the Mandelstam variables, $t$ isThe azimuthal integral contributes $2\pi$, and the angular endpoints are thereforeChanging variables gives the requested expression from the formula supplied in the paper:There is a normalization qualification: the displayed starting formula integrates over labeled final momenta and contains no $1/2!$. For the physical cross-section of two indistinguishable outgoing quanta of this real scalar field, the identical-particle factor in a final-state phase-space integral divides the full integral by two. With that convention $X=1/(32\pi )$; equivalently, integrate only one representative of each exchanged pair. The boxed value $1/(16\pi )$ follows the given formula exactly.

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.

With relativistically normalized external states, a two-particle initial state has differential cross-section $d\sigma =|\mathcal{M}{|}^{2}d{\Phi }_n/\mathcal{F}$, where $d{\Phi }_n$ is the Lorentz-invariant phase-space measure and $\mathcal{F}$ is the invariant flux factor. For identical unobserved final particles, divide the full labeled phase-space integral by their permutation multiplicity.