= Spin-reversal cross-ratio for beta decay
{title2=$\epsilon=(\sqrt R-1)/(\sqrt R+1)$}
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.<The Reversal of Parity Law in Nuclear Physics by National Institute of Standards and Technology>[Apparatus explanation and December 27 notebook: fixed beta counter, reversed solenoid current, and warming]{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 are
$$
N_+^{\,+}=C_+\eta_+(1+\epsilon),\qquad
N_-^{\,+}=C_+\eta_-(1-\epsilon),\qquad
N_+^{\,-}=C_-\eta_+(1-\epsilon),\qquad
N_-^{\,-}=C_-\eta_-(1+\epsilon),\qquad
\epsilon=ak,\quad |\epsilon|<1.
$$
The same-state ratios each retain the unequal <detection efficiencies>:
$$
\frac{N_+^{\,+}}{N_-^{\,+}}=\frac{\eta_+}{\eta_-}\frac{1+\epsilon}{1-\epsilon},\qquad
\frac{N_+^{\,-}}{N_-^{\,-}}=\frac{\eta_+}{\eta_-}\frac{1-\epsilon}{1+\epsilon}.
$$
Dividing these ratios gives the <spin-reversal cross-ratio for beta decay>:
$$
R=\frac{N_+^{\,+}N_-^{\,-}}{N_-^{\,+}N_+^{\,-}}
=\left(\frac{1+\epsilon}{1-\epsilon}\right)^2,\qquad
\epsilon=\frac{\sqrt R-1}{\sqrt R+1}.
$$
Both the exposure ratio $C_+/C_-$ and the <detection efficiency> ratio $\eta_+/\eta_-$ cancel. The positive square root is essential: positive expected counts and $|\epsilon|<1$ make $(1+\epsilon)/(1-\epsilon)$ positive.
For the chosen $a=-0.3$, opposite hemispheres give $k=1/2$, hence $\epsilon=-0.15$. 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 $3400$. Thus
$$
\frac{N_+^{\,+}}{N_-^{\,+}}=\frac{136}{115},\qquad
\frac{N_+^{\,-}}{N_-^{\,-}}=\frac{184}{85},\qquad
R=\frac{289}{529},\qquad \sqrt R=\frac{17}{23},\qquad
\epsilon=-\frac3{20}.
$$
The first raw detector ratio exceeds one despite negative $\epsilon$: detector imbalance can overwhelm the physical directional preference. Reversal separates the stable imbalance from the term that changes sign.
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