= Solution
Assume unpolarized beams, negligible <electron> and <muon> <masses>, a single <scattering s-channel> spin-one mediator with equal left and right vector couplings, and leading-order <QED>. <Z boson> exchange and radiative corrections are omitted. The angular factors are the <Wigner D-matrix> elements $d^1_{mm'}(\theta^\star)$, with $m,m'=\pm1$. Up to common phases and normalization they are $(1+\cos\theta^\star)$ for equal pair-spin projections and $(1-\cos\theta^\star)$ for opposite projections. The four orthogonal <helicity> channels add incoherently, giving
$$
2(1+\cos\theta^\star)^2+2(1-\cos\theta^\star)^2=4(1+\cos^2\theta^\star).
$$
With $s$ the squared <center-of-mass energy>, the normalized <differential scattering cross-sections> are
$$
\boxed{\frac{d\sigma}{d\Omega}=\frac{\alpha^2}{4s}(1+\cos^2\theta^\star),\qquad
\frac{d\sigma}{d\theta^\star}=\frac{\pi\alpha^2}{2s}\sin\theta^\star(1+\cos^2\theta^\star).}
$$
The second expression includes the <solid angle> Jacobian: a distribution per angle is different from one per $\cos\theta^\star$. Integration gives $\sigma=4\pi\alpha^2/(3s)$.
<Parity> reverses all <momenta> but leaves the <spin> axial vector unchanged, so it interchanges $L\leftrightarrow R$. The $LR\to LR$ and $RL\to RL$ amplitudes have equal magnitudes, as do $LR\to RL$ and $RL\to LR$. Thus every configuration has the same rate as its parity transform, proving <parity conservation>. The summed distribution is also forward-backward symmetric. Strictly, <parity> reverses both the beam and outgoing <momenta>, leaving their relative angle unchanged; forward-backward reflection by itself is not the definition of <parity>.
Back to article page
