= Solution
A representative <Feynman diagram> annihilates $u\bar d$ into $W^{+*}$, producing $\nu_\mu$ and an off-shell $\mu^+$; the charged <antilepton> emits a <photon> and becomes the outgoing $\mu^+$. The top panel below shows this contribution. A complete <gauge-invariant> <decay amplitude> includes radiation from the charged hadronic side, together with structure-dependent terms. The physical separation into inner <bremsstrahlung>, structure-dependent radiation, and their <interference> is explicit in the standard <radiative pion decay> description.<Radiative pion and kaon decay form factors in the 2025 Review of Particle Physics>[Section 62.1, equations (62.1)–(62.4): decomposition into inner bremsstrahlung, structure-dependent radiation and interference]{ref}
\Image[/past-exam-of-the-nuclear-and-particle-physics-course-of-the-university-of-oxford/2022-radiative-weak-decays.png]
{title=Representative radiative pion decay and internal-conversion muon decay diagrams}
Here is an explicit finite \b[model of <phase space> alone], with its assumptions exposed. Neglect the <neutrino> <mass>, treat the <photon> as massless, and replace the spin-summed invariant <decay amplitudes> by constants over the available <Lorentz-invariant phase space>. Write $M=m_\pi$, $m=m_\mu$, $r=m^2/M^2$, and use <natural units>. The <three-body phase space with one massive daughter> follows by factoring through an intermediate invariant squared <mass> $s$:
$$
\Phi_3(M;m,0,0)=\frac{1}{128\pi^3M^2}
\int_{m^2}^{M^2}\frac{(M^2-s)(s-m^2)}{s}\,ds
=\frac{M^2}{256\pi^3}(1-r^2+2r\ln r).
$$
The two-body comparison has $\Phi_2(M;m,0)=(1-r)/(8\pi)$. Since $\Gamma_n=\overline{|\mathcal M_n|^2}\Phi_n/(2M)$, assuming $\overline{|\mathcal M_3|^2}/\overline{|\mathcal M_2|^2}=\kappa e^2/M^2$ gives
$$
\boxed{\frac{\Gamma_{\mu\nu\gamma}}{\Gamma_{\mu\nu}}
\simeq\frac{\kappa\alpha_{\mathrm{em}}}{8\pi}
\frac{1-r^2+2r\ln r}{1-r}.}
$$
Here $\kappa$ is a dimensionless constant encapsulating the assumed <matrix element> ratio, and $\alpha_{\mathrm{em}}$ is the <fine-structure constant>. If the two-body mode dominates, this ratio approximates the radiative <branching fraction>. Near $m=M$, $1-r^2+2r\ln r=(1-r)^3/3+O((1-r)^4)$, displaying the extra <phase space> suppression.
\b[This model is not a prediction of a physical uncut radiative branching fraction.] Actual inner <bremsstrahlung> has a soft-<photon> singularity, so a count of events with a separately observed <photon> requires a positive <photon> <energy> cut. To see both the missing input and the <mass> dependence explicitly, the leading soft-<photon> result for $E_{\mathrm{cut}}<E_\gamma<E_{\mathrm{soft}}\ll E_{\max}$ is
$$
\boxed{\frac{\Gamma_{\gamma,\mathrm{soft}}}{\Gamma_{\mu\nu}}
\simeq\frac{\alpha_{\mathrm{em}}}{\pi}
\left[\frac{1+r}{1-r}\ln\frac1r-2\right]
\ln\frac{E_{\mathrm{soft}}}{E_{\mathrm{cut}}},\quad
E_{\max}=\frac{M^2-m^2}{2M}.}
$$
The coefficient follows from the radiation factor $e[p_\mu\cdot\epsilon/(p_\mu\cdot k)-p_\pi\cdot\epsilon/(p_\pi\cdot k)]$: angular integration gives $\beta^{-1}\ln[(1+\beta)/(1-\beta)]-2$, where $\beta=(1-r)/(1+r)$, and the soft <energy> integral is $\int dE_\gamma/E_\gamma$. Thus a finite physical number cannot be deduced from the two <masses> and <phase space> alone. The <infrared divergence> cancels only when the corresponding unresolved real radiation and virtual <radiative corrections> are combined in an inclusive observable.
Back to article page
