 Polarized beta-decay angular distribution

Consider a parent atomic nucleus at rest. Write its signed nuclear polarization along a fixed axis $\hat{\mathbf{z}}$ as $P=⟨J_z⟩/(I\hslash )$, where $I\hslash$ 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 $\mathbf{P}=P\hat{\mathbf{z}}$. Rotational symmetry makes the directional term proportional to $\mathbf{P}\cdot \hat{\mathbf{p}}$. Define its dimensionless coefficient as $A\beta$. At fixed Electron energy the relative intensity and normalized probability density function per solid angle are thereforeThe 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:Thus $f$ has unit angular average, while the probability density function in the polar angle alone contains the Jacobian determinant factor:In particular, $f$ itself is neither a normalized probability density function in $\theta$ nor a detector count. Positivity on $-1\le u\le 1$ gives $|a|\le 1$.

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 $\mathbf{P}\cdot \hat{\mathbf{p}}$ 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: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.