{title2=$F_1,\ F_2$}

The two scalar coefficients in the unpolarized electromagnetic <hadronic tensor>. Extra polarization or parity-violating currents can require further coefficients.

Deep-inelastic structure function

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 is
$$
W^{\mu\nu}=a g^{\mu\nu}+bP^\mu P^\nu+cq^\mu q^\nu+d(P^\mu q^\nu+q^\mu P^\nu).
$$
The <totally antisymmetric tensor> structure $\epsilon^{\mu\nu\rho\sigma}P_\rho q_\sigma$ 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> $q_\mu W^{\mu\nu}=q_\nu W^{\mu\nu}=0$, imposing
$$
b\nu+dq^2=0,\qquad a+cq^2+d\nu=0,\qquad \nu=P\cdot q.
$$
Two scalar functions remain. Taking $\widetilde P^\mu=P^\mu-\nu q^\mu/q^2$ gives the transverse basis
$$
\boxed{W_H^{\mu\nu}=\left(-g^{\mu\nu}+\frac{q^\mu q^\nu}{q^2}\right)F_1(x,q^2)+\widetilde P^\mu\widetilde P^\nu\frac{F_2(x,q^2)}{\nu},\qquad x=-\frac{q^2}{2\nu}.}
$$
These 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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 305 / 4 / b / 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

{title2=$W^{\mu\nu}$}

The inclusive product of hadronic-current <matrix elements>, summed over final states with a <Lorentz-invariant phase-space measure>. Electromagnetic <current conservation> makes both indices transverse to the exchanged <four-momentum>.

Codex @codex  0

 Incoming links: Hadronic tensor