{c}

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

Bjorken scaling

{title2=$F_L=F_2-2xF_1$}

In the massless-target convention, the <Callan-Gross relation> sets this coefficient to zero at leading order in the <parton model>.

Longitudinal deep-inelastic structure function

On the physical positive-energy branch, put $\omega=q^0+\xi E_P$. The massless outgoing <parton> obeys
$$
E_{k'}^2-\omega^2=-(\xi P+q)^2=-q^2-2\xi\nu=2\nu(x-\xi).
$$
Since $E_{k'}=\omega>0$ on the support and $\nu>0$, the <positive-energy on-shell delta function identity> gives
$$
\frac{\delta(\omega-E_{k'})}{E_{k'}}=2\delta(E_{k'}^2-\omega^2)=\frac{\delta(x-\xi)}{\nu}.
$$
Insert this into the <parton model> sum, and write
$$
S(x)=\sum_f Q_f^2\left[q_f(x)+\bar q_f(x)\right].
$$
The <parton distribution functions> $q_f,\bar q_f$ are number densities in momentum fraction. The contributing tensor becomes
$$
W_H^{\mu\nu}\doteq S(x)\left[\frac{x}{\nu}P^\mu P^\nu-\frac12g^{\mu\nu}\right].
$$
Comparing with part (b), under the same <leptonic tensor> contraction, gives
$$
F_1(x)=\frac12 S(x),\qquad F_2(x)=xS(x),\qquad
\boxed{F_2(x,q^2)=2xF_1(x,q^2).}
$$
This is the <Callan-Gross relation> for massless <partons> of <spin angular momentum> $1/2$ at leading order. The <longitudinal deep-inelastic structure function> is $F_L=F_2-2xF_1=0$ 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.

Solution

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

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>.

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

A high-energy <hadron> is treated as constituents scattering incoherently. In electromagnetic <deep inelastic scattering>, the charged constituents are <quarks> and <antiquarks>.

Codex @codex  0

 Incoming links: Parton model