= Solution
Assume the same left and right couplings $g_l,g_r$ at the <electron> and <muon> vertices. This <lepton universality> assumption is necessary to describe both vertices using just two numbers. Define $a=|g_l|^2$, $b=|g_r|^2$ and $z=\cos\theta^\star$. The two same-chirality channels have squared coupling weights $a^2,b^2$ and angular factor $(1+z)^2$; the other two each have weight $ab$ and factor $(1-z)^2$. Hence, for a common positive normalization $C$,
$$
\frac{d\sigma}{dz}=C\big[(a^2+b^2)(1+z)^2+2ab(1-z)^2\big]
=\boxed{C\big[(a+b)^2(1+z^2)+2(a-b)^2z\big].}
$$
If $|g_l/g_r|\ne1$, then $a\ne b$, so a forward-backward term survives. At the amplitude level, <parity> interchanges the $g_l^2$ and $g_r^2$ channels; their unequal squared magnitudes show directly that the interaction violates <parity>. Equivalently the chiral current contains both <vector current> and <axial current> terms, with coefficients $(g_r+g_l)/2$ and $(g_r-g_l)/2$; unequal magnitudes preclude a pure parity eigen-current.
The odd-in-$z$ term is a useful signature within these assumptions, but it should not itself be called a parity-odd scalar: the unpolarized relative angle is unchanged when \i[all] <momenta> are reversed. If the two species have independent couplings, the coefficient is instead proportional to $(a_e-b_e)(a_\mu-b_\mu)$, and a parity-violating coupling at only one vertex need not produce an unpolarized forward-backward asymmetry.
Back to article page
