{c}
{title2=$x=\frac{Q^2}{2P\cdot q}$}

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

Bjorken x

Use the <massless collinear parton approximation> in a high-energy frame: $P^2=0$, $k=\xi P$, and $k'=\xi P+q$ with $k'^2=0$. 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 $Q_f\bar u(k')\gamma^\mu u(k)$. The <spin average> and <gamma-matrix trace> give
$$
\frac12\sum_{\rm spins}J^\mu J^{\nu *}=2Q_f^2\left(k^\mu k'^\nu+k^\nu k'^\mu-g^{\mu\nu}k\cdot k'\right).
$$
Integrating the three-momentum <Dirac delta function> in the <parton> <hadronic tensor> leaves
$$
\widetilde W^{\mu\nu}=\frac{\delta(q^0+\xi E_P-E_{k'})}{2\xi E_{k'}}\left(k^\mu k'^\nu+k^\nu k'^\mu-g^{\mu\nu}k\cdot k'\right).
$$
Since $k\cdot k'=\xi\nu$, this is
$$
\widetilde W^{\mu\nu}=\frac{\delta(q^0+\xi E_P-E_{k'})}{E_{k'}}\left[\xi P^\mu P^\nu+\frac12(P^\mu q^\nu+q^\mu P^\nu)-\frac\nu2g^{\mu\nu}\right].
$$
For the massless <Electron> momenta, $p^2=p'^2=0$ and $q=p-p'$. Substitution into the <leptonic tensor> gives
$$
q^\mu L_{\mu\nu}=4\left[-(p\cdot p')p'_\nu+(p\cdot p')p_\nu-(p\cdot p')(p_\nu-p'_\nu)\right]=0,
$$
and likewise $q^\nu L_{\mu\nu}=0$. These <Ward identities> eliminate every term with an exposed $q$ index in the contraction. Therefore
$$
\boxed{\widetilde W^{\mu\nu}\doteq\frac{\delta(q^0+\xi E_P-E_{k'})}{E_{k'}}\left[\xi P^\mu P^\nu-\frac{\nu}{2}g^{\mu\nu}\right].}
$$
Here $\doteq$ means equality \b[after contraction with the <leptonic tensor>]. The shortened tensor is not itself conserved; the omitted terms restore <current conservation> in the full <hadronic tensor>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 305 / 4 / c / i / Solution

{title2=$\frac{\delta(\omega-E)}{E}=2\delta(E^2-\omega^2),\quad \omega>0$}

For $E>0$ and the positive-energy branch, the root Jacobian of the <Dirac delta function> gives this identity. In a <massless collinear parton approximation>, it produces $\delta(x-\xi)/(P\cdot q)$.

Positive-energy on-shell delta function identity

{title2=$k=\xi P,\quad P^2=k^2=0$}

The leading high-energy approximation neglects <quark> and target masses and transverse constituent momentum. A lightlike reference momentum is used for the <hadron>; this does not describe a massless particle at rest.

Codex @codex  0

 Incoming links: Massless collinear parton approximation