= Polarized beta-decay angular distribution
{title2=$f(\theta)=1+A P\beta\cos\theta$}
Consider a parent <atomic nucleus> at rest. Write its signed <nuclear polarization> along a fixed axis $\widehat{\mathbf z}$ as $P=\langle J_z\rangle/(I\hbar)$, where $I\hbar$ is its maximal <spin> component. Let $\theta$ be the angle of the emitted <Electron> <momentum> to that axis and let $\beta=v/c$. Use $\mathcal P$ for the <parity operator>, so it is not confused with $P$.
For an axially symmetric leading-dipole model, retain a constant and the term linear in the polarization <vector> $\mathbf P=P\widehat{\mathbf z}$. <Rotational symmetry> makes the directional term proportional to $\mathbf P\cdot\widehat{\mathbf p}$. Define its dimensionless coefficient as $A\beta$. At fixed <Electron> <energy> the relative intensity and normalized <probability density function> per <solid angle> are therefore
$$
f(\theta)=1+a\cos\theta,\qquad a=A P\beta,\qquad
\frac{d\Pr}{d\Omega}=\frac{1+a\cos\theta}{4\pi},\qquad |a|\leq1.
$$
The factor $\beta$ is the conventional separation of the speed dependence; <rotational symmetry> alone does not determine that dependence or $A$. The coefficient $A$ contains the decay dynamics. This leading model omits higher angular terms and detailed recoil and radiative effects. A partially oriented higher-<spin> ensemble can also have alignment that is not specified by its vector <nuclear polarization> alone.
The unit-integral condition follows without any model of a detector:
$$
\int d\Omega=4\pi,\qquad
\int \cos\theta\,d\Omega=2\pi\int_{-1}^{1}u\,du=0.
$$
Thus $f$ has unit angular average, while the <probability density function> in the polar angle alone contains the <Jacobian determinant> factor:
$$
\frac{d\Pr}{d\theta}=\frac12(1+a\cos\theta)\sin\theta,\qquad
\frac{d\Pr}{du}=\frac12(1+au),\quad u=\cos\theta.
$$
In particular, $f$ itself is neither a normalized <probability density function> in $\theta$ nor a detector count. Positivity on $-1\leq u\leq1$ gives $|a|\leq1$.
Under full <spatial reflection>, $\mathbf p\mapsto-\mathbf p$ but $\mathbf J\mapsto\mathbf J$. The latter follows already for <orbital angular momentum> from $\mathbf r\times\mathbf p\mapsto(-\mathbf r)\times(-\mathbf p)=\mathbf r\times\mathbf p$; <spin> has the same <axial vector> transformation. Hence $\mathbf P\cdot\widehat{\mathbf p}$ is a <pseudoscalar>, and $\cos\theta\mapsto-\cos\theta$. If <parity conservation> held for this model, equal <probabilities> for the inverted process would force $a=0$. Conversely, $a=0$ in an unpolarized sample proves nothing about $A$.
Integrating over the two hemispheres gives a useful original check:
$$
\Pr(u>0)=\frac12+\frac a4,\qquad
\Pr(u<0)=\frac12-\frac a4,\qquad
\mathcal A_{\rm hemispheres}=\frac a2.
$$
Choose $A=-0.8$, $P=0.5$, and $\beta=0.75$, so $a=-0.3$. Then the forward and backward hemisphere <probabilities> are $0.425$ and $0.575$, and their counting asymmetry is $-0.15$. These are chosen teaching values, not measured values from the <Wu experiment>. At a single <energy>, the angular shape determines the product $AP\beta$; it cannot separately determine $A$ without a polarization calibration.
\Image[/polarized-beta-decay-angular-law-and-cross-ratio.png]
{title=Chosen angular probabilities and detector-efficiency cancellation}
{description=Original model with $a=-0.3$. The angular probability is distinct from a detector count; opposite hemispheres average the cosine to $\pm1/2$. Unequal efficiencies and exposures distort individual detector ratios, while a reversal cross-ratio recovers the chosen hemispheric asymmetry. All counts are constructed.}
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