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A detector collects a range of solid angles, not a single direction. If its nonnegative angular response is $w(\Omega )$, defineIntegrating the polarized beta-decay angular distribution gives a count proportional to $G(1+ak)$. For a uniform forward cone with half-angle ${\theta }_0$, put $c_0=\cos {\theta }_0$. Direct integration givesThe opposite cone has mean cosine $-k$. A hemisphere gives $k=1/2$; a narrow cone tends to $k=1$. Hence a spin-reversal cross-ratio for beta decay measures $ak$, and an acceptance correction is needed to obtain $a$.

With unequal angular responses, let the two laboratory mean cosines be $k_{+}$ and $k_{-}$, including their signs. Then the actual expectation iswhich reduces to the simple square only when $k_{-}=-k_{+}$. The cancellation of stable multiplicative detection efficiencies survives, but their angular shapes still matter. An energy-integrated response likewise averages $A(E)P\beta (E)\cos \theta$ with the accepted spectrum; a mean speed cannot in general be substituted independently of acceptance.

Additive backgrounds, unequal polarization magnitudes or detection efficiencies that change on reversal do not have the assumed factorization. For example, a reversal-dependent efficiency cross-ratio multiplies $R$ by ${\eta }_{+}^{+}{\eta }_{-}^{-}/({\eta }_{-}^{+}{\eta }_{+}^{-})$. Background subtraction also creates statistical complications: positive expected signal counts are assumed in this derivation, whereas finite samples with Poisson distributions can contain zero counts. Taking square roots of negative background-subtracted counts is not a valid estimator. These limits explain why reversal and independent preparation checks belong together.

Consider a teaching extension with two opposite detectors, labelled $+$ and $-$ by fixed laboratory directions, and two preparation states $s=+1,-1$ with equal and opposite nuclear polarization. This is an original two-detector model; the historical Wu experiment used a fixed internal Electron counter with reversal.^[ref] Let $C_s>0$ describe the exposure in each state and ${\eta }_d>0$ the stable detection efficiency of detector $d$.

For matched opposite acceptances, write their common mean absolute cosine as $k$. The four expected positive counts areThe same-state ratios each retain the unequal detection efficiencies:Dividing these ratios gives the spin-reversal cross-ratio for beta decay:Both the exposure ratio $C_{+}/C_{-}$ and the detection efficiency ratio ${\eta }_{+}/{\eta }_{-}$ cancel. The positive square root is essential: positive expected counts and $|ϵ|<1$ make $(1+ϵ)/(1-ϵ)$ positive.

For the chosen $a=-0.3$, opposite hemispheres give $k=1/2$, hence $ϵ=-0.15$. Choose $C_{+}=10000$, $C_{-}=8000$, ${\eta }_{+}=0.8$, and ${\eta }_{-}=0.5$, with their common acceptance factor absorbed into $C_s$. The four expected counts in the displayed order are $6800$, $5750$, $7360$, and $3400$. ThusThe first raw detector ratio exceeds one despite negative $ϵ$: detector imbalance can overwhelm the physical directional preference. Reversal separates the stable imbalance from the term that changes sign.