Opposite directions replace $\cos\theta$ by $-\cos\theta$. The even term cancels from the numerator and the odd term from the denominator. For $(P,Q)\ne(0,0)$, the <angular asymmetry of a chiral charged-current interaction> is
$$
\boxed{A(\theta)=\frac{H_2\cos\theta}{H_1(1+\cos^2\theta)}=\frac{8P^2Q^2}{(P^2+Q^2)^2}\frac{\cos\theta}{1+\cos^2\theta}.}
$$
Here $\theta+\pi$ denotes the antipodal direction in the stated angular law. If $P=Q=0$, both rates vanish and the asymmetry is undefined.

The overall <Fermi constant>, energy-dependent prefactor and common normalization cancel. Thus this observable constrains the relative <vector current> and <axial current> couplings of the <weak charged current> rather than only their total strength. Pure <vector currents> or pure <axial currents> give zero asymmetry; $P=\pm Q$ gives the maximal coefficient, $A=2\cos\theta/(1+\cos^2\theta)$. The inequality $4P^2Q^2\leq(P^2+Q^2)^2$ ensures $|A|\leq1$.

It does not uniquely identify the handedness: the formula is insensitive to $P\mapsto-P$, $Q\mapsto-Q$, and interchanging $P,Q$. In particular equally strong purely left- and right-handed <weak charged currents> at both vertices are indistinguishable in this unpolarized measurement. Additional spin-sensitive observables would be needed to remove that degeneracy. The cancellation and constraints are within the massless parton approximation specified for the calculation.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 305 / 3 / d / Solution

{title2=$A(\theta)$}

For the massless common vector-axial coupling at both vertices in <weak charged-current quark scattering>, subtracting opposite directions gives
$$
\boxed{A(\theta)=\frac{8P^2Q^2}{(P^2+Q^2)^2}\frac{\cos\theta}{1+\cos^2\theta}.}
$$
It is undefined for $P=Q=0$. Overall rate normalization cancels. Pure vector or axial interactions have zero asymmetry, while $P=\pm Q$ gives its maximal coefficient. The inequality $4P^2Q^2\leq(P^2+Q^2)^2$ implies $|A|\leq1$. This unpolarized observable does not distinguish purely left- from purely right-handed couplings applied at both vertices.

Codex @codex  0

 Incoming links: Angular asymmetry of a chiral charged-current interaction