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Take metric signature $(+,-,-,-)$ and the supplied covariant electromagnetic field tensor, so $F_{0i}=E_i$ and $F_{ij}=-{ϵ}_{ijk}{B}_k$. Raising both indices reverses the time-space components but leaves the spatial components unchanged. Its complete contraction is thereforeDefine the dual electromagnetic field tensor by ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{\epsilon }^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$, with ${\epsilon }^{0123}=+1$. Direct contraction givesThese electromagnetic field invariants are complete tensor contractions. The second is a pseudoscalar: it is invariant under proper Lorentz transformations and changes sign under spatial inversion.

For $T^{0i}$ the metric trace term vanishes. Since $F^0{}_j=E_j$,Consequently the Poynting vector is exactly the time-space row in the stated $c=1$ units:

The scalar contraction of the electromagnetic field tensor isWith orientation ${ϵ}^{0123}=+1$, define the dual electromagnetic field tensor by ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. Direct substitution givesThe first is a Lorentz scalar; the second is a pseudoscalar, invariant under proper Lorentz transformations such as boosts and rotations. Under spatial inversion $\mathbf{E}$ changes sign while $\mathbf{B}$ does not, so $\mathbf{E}\cdot \mathbf{B}$ reverses sign. For example, $\mathbf{E}=\mathbf{B}=\hat{\mathbf{z}}$ has product $+1$ before inversion and $-1$ afterwards. Thus the unrestricted statement including improper transformations needs this qualification.

For the spacetime integral, assume a globally regular electromagnetic four-potential on ordinary Minkowski space and sufficient boundary decay. Using $\nabla \cdot \mathbf{B}=0$ and ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$,ConsequentlyThe magnetic helicity boundary term controls the integral. If its surface flux vanishes and the magnetic helicity $H_B(t)=\int \mathbf{A}\cdot \mathbf{B}{d}^{3}x$ vanishes at both temporal ends, thenCompactly supported smooth potentials are one sufficient version of the intended falloff assumption. The local identity is exact, but vanishing fields at spatial infinity alone is insufficient. For a concrete counterexample, take $\varphi =0$ and, in dimensionless coordinates,where $a$ changes smoothly from zero to one and $\dot{a}$ has compact temporal support. Both electric fields and magnetic fields vanish rapidly at spatial infinity, butThe counterexample leaves a final magnetic field and therefore fails the extra temporal boundary condition. It shows exactly why the printed integral cannot follow from a merely spatial reading of “at infinity”; vanishing of the full boundary term is the needed hypothesis.

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.

Between spin-zero pseudoscalars, Lorentz covariance decomposes a vector current matrix element into $(p+k{)}^{\mu }{f}_{+}(q^2)+(p-k{)}^{\mu }{f}_{-}(q^2)$, with $q=p-k$. Strong interaction parity symmetry in quantum field theory makes the corresponding axial current matrix element vanish.

A pseudoscalar Yukawa interaction couples a Dirac field bilinear containing the chirality matrix to a spin-zero field. In the standard mostly-minus gamma matrix convention with ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$ and the usual Dirac adjoint, $ig\bar{\psi }{\gamma }^5\psi \varphi$ is Hermitian for real $g$ and real $\varphi$. If $\varphi$ is a pseudoscalar, the interaction also preserves parity symmetry in quantum field theory; if $\varphi$ is parity even, it breaks that symmetry. Moving the factor $i$ into the definition of ${\gamma }^5$ changes the displayed vertex coefficient but not the interaction. Each vertex has one bosonic leg and two oppositely oriented fermion legs, and preserves Dirac fermion number conservation.

A linear scalar counterterm fixes the one-point function or the expansion point of the scalar potential. An unbroken parity symmetry in quantum field theory forbids this term when the scalar is a pseudoscalar. A vanishing one-loop tadpole diagram alone would not prove the all-orders restriction without the symmetry argument.