Muon decay

Muon decay, $\mu^-\to e^-+\bar\nu_e+\nu_\mu$ , is the cleanest laboratory the weak interaction offers. A muon lives for $2.2$ microseconds and then converts to a muon-neutrino by emitting a virtual $W^-$ ; the $W^-$ immediately materialises as an electron and an electron-antineutrino. There are no hadrons anywhere, so unlike nuclear beta decay there is no strong-interaction structure to model — every ingredient follows from the same Feynman rules as QED, changed only by a massive mediator and a chiral $V-A$ vertex factor $\tfrac{-ig_W}{2\sqrt2}\gamma^\mu(1-\gamma^5)$ . This is why muon decay is the reference process against which the charged-current weak interaction is calibrated. Two facts make the calculation tractable and the result famous. First, the momentum transferred to the $W$ is a few tens of MeV while $M_W\approx 80$ GeV, so the propagator collapses to a constant and the exchange becomes Fermi's four-fermion contact interaction with a single strength $G_F$ . Second, $g_W$ and $M_W$ enter only through the combination $G_F/\sqrt2 = g_W^2/8M_W^2$ , so measuring the muon lifetime measures $G_F$ directly, and to spectacular precision. Carrying the three-body phase-space integral through gives the Michel electron spectrum and the celebrated total rate $\Gamma = G_F^2 m_\mu^5/192\pi^3$ , the fifth-power law tying a particle's mass to how fast it decays. This article assembles that result from the $V-A$ vertex and the trace technology, and shows why the answer is trusted to parts per million.

Key takeaways

Muon decay is a single charged-current $V-A$ vertex used twice, joined by a $W$ propagator: $\mu^-\to \nu_\mu + W^{-*}\to \nu_\mu + e^- + \bar\nu_e$ .
At low momentum transfer $q^2\ll M_W^2$ the $W$ propagator becomes a constant, recovering Fermi's contact interaction with $G_F/\sqrt2 = g_W^2/8M_W^2$ .
The spin-averaged matrix element is $\langle|\mathcal M|^2\rangle = 64\,G_F^2\,(p_\mu\!\cdot p_{\bar\nu_e})(p_e\!\cdot p_{\nu_\mu})$ , a two-line trace result.
The electron energy spectrum is the Michel spectrum $\tfrac{d\Gamma}{dx}\propto x^2(3-2x)$ with $x=2E_e/m_\mu$ , ending at $E_{\max}=m_\mu/2\approx 52.8$ MeV.
The total rate is $\Gamma = G_F^2 m_\mu^5/192\pi^3$ , giving $\tau_\mu\approx 2.2\ \mu$ s and fixing $G_F=1.1664\times10^{-5}\,\mathrm{GeV}^{-2}$ .
The intrinsic weak coupling $\alpha_W=g_W^2/4\pi\approx 1/30$ is larger than the QED $\alpha=1/137$ ; the weakness is purely the heavy- $W$ propagator.
Key idea: muon decay is weak interactions in their simplest form — one $V-A$ vertex twice over, contracted through a heavy $W$ propagator that collapses to a contact interaction, so the whole rate depends only on $G_F$ and the muon mass to the fifth power.

The core insight

Muon decay is weak interactions in their simplest form: a single $V-A$ vertex twice over, contracted through a heavy $W$ propagator that collapses to a contact interaction, so the whole rate depends only on $G_F$ and the muon mass to the fifth power.

Where this lives today

Muon decay is not a settled historical calculation — it is a live precision instrument. The MuLan experiment at PSI measured the muon lifetime to $\tau_\mu=2\,196\,980.3\pm2.2$ ps, a $1$ -ppm result that fixes $G_F=1.1663787(6)\times10^{-5}\,\mathrm{GeV}^{-2}$ , one of the best-known constants in physics and a required input to every electroweak precision fit. The TWIST experiment at TRIUMF measured the Michel parameters $\rho,\delta,\xi$ to the fourth decimal, tightening bounds on any non- $V-A$ (scalar, tensor, right-handed) admixture and on $W_R$ bosons in left-right symmetric models. And the forbidden versions of the decay are among the sharpest probes of new physics: MEG-II hunts for $\mu^+\to e^+\gamma$ (branching limit below $10^{-13}$ ), Mu3e searches for $\mu^+\to e^+e^-e^+$ , and Mu2e and COMET look for coherent $\mu\to e$ conversion in nuclei — all charged-lepton-flavour violating, where a single event would be unambiguous new physics. The same vertex, run at high energy, is what a $W$ does when it decays leptonically at the LHC: the contact interaction resolved into its full propagator form.

The physical picture

A muon at rest carries nowhere to go but downhill in mass. The only lighter charged lepton is the electron, and lepton-family numbers must balance, so the muon sheds its muon-ness into a $\nu_\mu$ and hands the rest to an $e^-$ accompanied by a $\bar\nu_e$ that keeps electron-number zero. The transaction is brokered by a single virtual $W^-$ : the muon touches the $W$ at one $V-A$ vertex and the electron pair touches it at another. Because the $W$ is thirty times heavier than the muon it can never be real here — it is borrowed for an instant and its heavy mass is exactly what makes the whole thing look like a point-contact interaction. The first figure is that diagram, with the vertex factor and the coupling relation that governs it.

Figure

The muon decay Feynman diagram: the μ⁻ emits a virtual W⁻ and becomes a ν_μ, and the W⁻ materialises as e⁻ and ν̄_e. Each vertex carries the V−A factor −ig_W/(2√2) γ^μ(1−γ⁵), and only the ratio g_W/M_W survives as the Fermi constant, with α_W = g_W²/4π ≈ 1/30 larger than the QED α = 1/137

The observable that turns this cartoon into physics is the electron's energy. The muon at rest shares its rest energy among three bodies, so the electron does not come out with a single energy but with a continuous spectrum — and the precise shape of that spectrum is a fingerprint of the interaction's chiral structure. Pure $V-A$ predicts one specific curve; a scalar or tensor admixture would bend it. The second figure is that spectrum, the Michel spectrum, with the numbers a real experiment reports.

Figure

The Michel spectrum: the electron energy distribution 2x²(3−2x) with x = 2E_e/m_μ, rising to a sharp endpoint at E_max = m_μ/2 = 52.8 MeV. Its shape parameter is the Michel ρ = 3/4, the pure V−A prediction, measured as ρ = 0.74977 ± 0.00012

The core equations

The starting point is the amplitude written with the physical $W$ still in place. Two $V-A$ vertices sandwich the $W$ propagator, one current carrying the muon into its neutrino, the other creating the electron pair. The Lorentz index of the propagator ties the two currents together.

\mathcal M = \frac{g_W^2}{8}\,\big[\bar u(\nu_\mu)\gamma^\mu(1-\gamma^5)u(\mu)\big]\,\frac{-g_{\mu\nu}+q_\mu q_\nu/M_W^2}{q^2-M_W^2}\,\big[\bar u(e)\gamma^\nu(1-\gamma^5)v(\bar\nu_e)\big] \qquad (1)

where: $g_W$ — weak charged-current coupling; $q$ — four-momentum carried by the $W$ ; $M_W$ — $W$ mass; $u,v$ — Dirac spinors; $(1-\gamma^5)$ — the left-chiral projector (twice the projector $P_L$ ) that makes the interaction $V-A$ . Reading it: this is structurally identical to a QED amplitude with photon exchange, but the vertex is chiral and the mediator is massive. Because the momentum transfer is at most tens of MeV while $M_W\approx 80\,\mathrm{GeV}$ , the $q^2$ and $q_\mu q_\nu$ terms are utterly negligible and the propagator collapses to the constant $g_{\mu\nu}/M_W^2$ . The exchange becomes a point contact between two currents, and the surviving strength is baptised the Fermi constant.

\frac{G_F}{\sqrt2}=\frac{g_W^2}{8M_W^2},\qquad \mathcal M=\frac{G_F}{\sqrt2}\big[\bar u(\nu_\mu)\gamma^\mu(1-\gamma^5)u(\mu)\big]\big[\bar u(e)\gamma_\mu(1-\gamma^5)v(\bar\nu_e)\big] \qquad (2)

where: $G_F=1.1663787(6)\times10^{-5}\,\mathrm{GeV}^{-2}$ — the Fermi constant; the $\sqrt2$ is a historical convention fixed so that this exact process defines $G_F$ . Reading it: $g_W$ and $M_W$ have merged into a single measurable number. Muon decay cannot separate the coupling from the mass — only their ratio. To get a rate we square the amplitude, average over the muon's spin and sum over the final spins. The neutrino chirality projectors do the bookkeeping, and the two Dirac traces collapse to a strikingly compact Lorentz-invariant product of dot products (electron mass neglected).

\langle|\mathcal M|^2\rangle = \frac{1}{2}\sum_{\text{spins}}|\mathcal M|^2 = 64\,G_F^2\,(p_\mu\!\cdot p_{\bar\nu_e})\,(p_e\!\cdot p_{\nu_\mu}) \qquad (3)

where: $p_i$ — four-momenta of the labelled particles; the factor $\tfrac12$ averages the two muon spin states. Reading it: the electron pairs with the muon-neutrino and the muon with the electron-antineutrino — a chirality pairing that is the whole imprint of $V-A$ on the rate. Integrating over the two unobserved neutrinos while holding the electron energy fixed yields the Michel spectrum: the probability that the electron emerges with reduced energy $x=2E_e/m_\mu$ (so $x\in[0,1]$ with the electron mass neglected).

\frac{d\Gamma}{dx}=\frac{G_F^2 m_\mu^5}{192\pi^3}\,\big[\,2x^2(3-2x)\,\big],\qquad x=\frac{2E_e}{m_\mu} \qquad (4)

where: the bracket integrates to unity over $x\in[0,1]$ , so the prefactor is the total rate; the shape $x^2(3-2x)$ is the pure $V-A$ prediction and its curvature is parametrised by the Michel $\rho=3/4$ . Reading it: the spectrum vanishes as $x^2$ at low energy and climbs to a maximum right at the endpoint. Integrating the spectrum over all electron energies gives the total width, and with it the lifetime — the fifth-power law that is the signature of this whole family of leptonic decays.

\Gamma=\frac{G_F^2 m_\mu^5}{192\pi^3},\qquad \tau_\mu=\frac{\hbar}{\Gamma} \qquad (5)

where: $m_\mu=105.6584$ MeV — muon mass; $\hbar=6.582\times10^{-25}\,\mathrm{GeV\,s}$ . Reading it: the rate scales as $m_\mu^5$ because five powers of energy are needed on dimensional grounds once $G_F^2$ contributes $\mathrm{GeV}^{-4}$ — three from three-body phase space and two from the matrix element. For parts-per-million work the leading formula needs two multiplicative corrections: the finite electron mass shrinks the phase space slightly, and QED radiative corrections (real and virtual photons) shift the rate by half a percent. Both are calculable.

\Gamma=\frac{G_F^2 m_\mu^5}{192\pi^3}\,f\!\left(\frac{m_e^2}{m_\mu^2}\right)\Big(1+\frac{\alpha}{2\pi}\Big(\tfrac{25}{4}-\pi^2\Big)+\dots\Big),\qquad f(y)=1-8y+8y^3-y^4-12y^2\ln y \qquad (6)

where: $f\approx0.99981$ is the electron-mass phase-space factor and the $\alpha$ term is $\approx0.9958$ . Reading it: these corrections are why the naive formula gives $2.187\ \mu$ s while the measured lifetime is $2.197\ \mu$ s — and why, run in reverse, muon decay yields $G_F$ to sub-ppm precision.

Derivation AFrom W exchange to the Fermi contact interaction

The goal is to show precisely why a process mediated by an 80 GeV boson looks, at muon energies, like Fermi's 1934 point interaction, and to pin down the exact relation between the fundamental coupling $g_W$ , the boson mass $M_W$ , and the phenomenological Fermi constant $G_F$ . We start from the propagator in amplitude (1), take the low-momentum-transfer limit, and match the resulting effective vertex onto the four-fermion form, keeping careful track of the numerical factors that fix the conventional $\sqrt2$ .

Step 1. Look at the $W$ propagator $\dfrac{-g_{\mu\nu}+q_\mu q_\nu/M_W^2}{q^2-M_W^2}$ . In muon decay the momentum $q$ transferred through the $W$ is bounded by the muon mass, so $|q^2|\lesssim m_\mu^2\approx(0.1\,\mathrm{GeV})^2$ , whereas $M_W^2\approx(80\,\mathrm{GeV})^2$ . The ratio $q^2/M_W^2\sim10^{-6}$ . Why: the propagator carries the entire energy dependence of the exchange, and this estimate is what justifies discarding it — the massive mediator is being probed a million times below its natural scale.

Step 2. Drop $q^2$ against $M_W^2$ in the denominator and drop $q_\mu q_\nu/M_W^2$ against $g_{\mu\nu}$ in the numerator (the latter term also vanishes when contracted with the light-fermion currents, by the Dirac equation). The propagator becomes the constant $g_{\mu\nu}/M_W^2$ . Why: a constant propagator is not a propagator at all — it is a contact term, so the two currents now touch at a single spacetime point exactly as in Fermi's original theory.

Step 3. Reassemble the amplitude with the two vertex factors $-ig_W/(2\sqrt2)\,\gamma^\mu(1-\gamma^5)$ and the constant propagator. The couplings collect into $g_W^2/(8M_W^2)$ multiplying the current-current product. Define this to be $G_F/\sqrt2$ , giving $G_F/\sqrt2 = g_W^2/8M_W^2$ . Why: this is the definition of $G_F$ — it packages the two things muon decay cannot separately measure, the coupling and the mediator mass, into the one thing it can.

Step 4. Check the scale. With $g_W\approx0.65$ and $M_W=80.4$ GeV, $G_F=\sqrt2\,g_W^2/8M_W^2\approx1.17\times10^{-5}\,\mathrm{GeV}^{-2}$ , matching the measured value. Equivalently $\alpha_W=g_W^2/4\pi\approx1/30$ , larger than $\alpha=1/137$ . Why: the numbers confirm the punchline — the coupling is not small, so the observed feebleness of the weak force at low energy is entirely the $1/M_W^2$ suppression sitting in $G_F$ .

Go deeper · Derivation B · 1 minThe spin-averaged matrix element by trace technology

Derivation BThe spin-averaged matrix element by trace technology

The aim is to turn the amplitude (2) into a number that depends only on the external momenta, by squaring it, averaging over the initial muon spin, summing over final spins, and evaluating the resulting Dirac traces. The payoff is the remarkably compact result (3), whose structure — electron paired with muon-neutrino — encodes the chirality of the weak interaction. We use completeness of spinors to convert spin sums into traces and the standard gamma-matrix trace identities.

Step 1. Write $|\mathcal M|^2 = (G_F^2/2)\,L^{(\mu)}_{\alpha\beta}\,L^{(e)\,\alpha\beta}$ , a product of a muon tensor and an electron tensor, each of the form $\sum \bar u\Gamma^\alpha u\,(\bar u\Gamma^\beta u)^*$ . Using $\sum u\bar u = (\slashed p+m)$ and $\sum v\bar v=(\slashed p-m)$ , each spin sum becomes a trace of gamma matrices. Why: spinor completeness is the device that removes all reference to individual spin states, replacing them with momenta — the only variables a rate can depend on.

Step 2. Evaluate the muon tensor $L^{(\mu)}_{\alpha\beta}=\mathrm{Tr}\big[(\slashed p_\mu+m_\mu)\gamma_\alpha(1-\gamma^5)\slashed p_{\nu_\mu}\gamma_\beta(1-\gamma^5)\big]$ . The $(1-\gamma^5)$ projectors kill every term with an odd number of gamma matrices, so the $m_\mu$ piece drops out. The surviving trace gives $8\big[p_{\mu\alpha}p_{\nu_\mu\beta}+p_{\mu\beta}p_{\nu_\mu\alpha}-g_{\alpha\beta}(p_\mu\!\cdot p_{\nu_\mu})-i\epsilon_{\alpha\beta\rho\sigma}p_\mu^\rho p_{\nu_\mu}^\sigma\big]$ . Why: the chiral projector is doing real physical work here — it removes the mass term, which is why the decaying muon's spin correlations survive and why the antisymmetric $\epsilon$ term (parity violation) appears at all.

Step 3. Contract the muon tensor with the identical-form electron tensor. The two symmetric parts contract to $2\big[(p_\mu\!\cdot p_e)(p_{\nu_\mu}\!\cdot p_{\bar\nu_e})+(p_\mu\!\cdot p_{\bar\nu_e})(p_{\nu_\mu}\!\cdot p_e)\big]$ , and the two antisymmetric $\epsilon$ terms contract to $2\big[(p_\mu\!\cdot p_e)(p_{\nu_\mu}\!\cdot p_{\bar\nu_e})-(p_\mu\!\cdot p_{\bar\nu_e})(p_{\nu_\mu}\!\cdot p_e)\big]$ . Why: adding the symmetric and antisymmetric contractions is where the magic happens — the $(p_\mu\!\cdot p_e)(p_{\nu_\mu}\!\cdot p_{\bar\nu_e})$ terms cancel and the others double.

Step 4. Collect: $L^{(\mu)}\!\cdot L^{(e)}=256\,(p_\mu\!\cdot p_{\bar\nu_e})(p_e\!\cdot p_{\nu_\mu})$ , so with the $G_F^2/2$ prefactor, $\langle|\mathcal M|^2\rangle=64\,G_F^2\,(p_\mu\!\cdot p_{\bar\nu_e})(p_e\!\cdot p_{\nu_\mu})$ . Why: this compact pairing is the fingerprint of $V-A$ — the muon talks to the antineutrino it did not emit, and the electron to the neutrino, a correlation no parity-conserving theory would produce.

Go deeper · Derivation C · 1 minPhase-space integration and the fifth-power law

Derivation CPhase-space integration and the fifth-power law

The objective is to integrate the matrix element (3) over the three-body final-state phase space and arrive at both the electron spectrum (4) and the total rate (5). The key technical move is to do the two-neutrino integral first, since the neutrinos are unobserved, which reduces the problem to a single integral over the electron energy. We work in the muon rest frame with the electron treated as massless.

Step 1. Write the decay rate as $\Gamma=\dfrac{1}{2m_\mu}\displaystyle\int \langle|\mathcal M|^2\rangle\,(2\pi)^4\delta^4(p_\mu-p_e-p_{\nu_\mu}-p_{\bar\nu_e})\,\widetilde{dp_e}\,\widetilde{dp_{\nu_\mu}}\,\widetilde{dp_{\bar\nu_e}}$ , where each $\widetilde{dp}=d^3p/((2\pi)^3 2E)$ is the Lorentz-invariant measure. Why: this is the master formula for any decay rate — a matrix element weighted by the invariant phase space and energy-momentum conservation, normalised by the parent energy.

Step 2. Do the neutrino pair integral. Define $q=p_\mu-p_e=p_{\nu_\mu}+p_{\bar\nu_e}$ , the total neutrino momentum. The tensor integral $\int \widetilde{dp_{\nu_\mu}}\widetilde{dp_{\bar\nu_e}}\,p_{\nu_\mu}^\alpha p_{\bar\nu_e}^\beta\,(2\pi)^4\delta^4(q-p_{\nu_\mu}-p_{\bar\nu_e})$ is fixed by Lorentz invariance to be $\tfrac{1}{96\pi}(q^2 g^{\alpha\beta}+2q^\alpha q^\beta)$ . Why: the neutrinos are never measured, so integrating them out first is both physically natural and mathematically decisive — it turns a many-variable integral into a standard invariant tensor.

Step 3. Contract with the rest of the matrix element and use $q^2=(p_\mu-p_e)^2=m_\mu^2-2m_\mu E_e$ in the rest frame. The result, expressed in $x=2E_e/m_\mu$ , is $\dfrac{d\Gamma}{dx}=\dfrac{G_F^2 m_\mu^5}{192\pi^3}\,2x^2(3-2x)$ — the Michel spectrum. Why: this is the shape the figure plots; the $x^2$ threshold behaviour is pure phase space and the $(3-2x)$ falloff is the dynamics, i.e. the $V-A$ structure.

Step 4. Integrate over $x$ from $0$ to $1$ : $\int_0^1 2x^2(3-2x)\,dx = \big[2x^3-x^4\big]_0^1 = 1$ , so $\Gamma=\dfrac{G_F^2 m_\mu^5}{192\pi^3}$ . Numerically $\Gamma=3.01\times10^{-19}$ GeV and $\tau_\mu=\hbar/\Gamma=2.19\ \mu$ s. Why: the spectrum was normalised so its integral is exactly the total rate, and the fifth-power dependence on $m_\mu$ falls straight out of the $\mathrm{GeV}^{-4}$ dimension of $G_F^2$ plus the phase-space powers — the celebrated Sargent-type law.

Limiting cases

Case	Result	Meaning
Low momentum transfer $q^2\ll M_W^2$	propagator $\to g_{\mu\nu}/M_W^2$ ; $G_F/\sqrt2=g_W^2/8M_W^2$	the Fermi contact interaction — the regime of all muon and beta decay
High energy $q^2\sim M_W^2$	full propagator needed; $G_F$ no longer constant	the effective theory breaks down; the $W$ can go on-shell (collider physics)
Massless electron $m_e\to0$	endpoint at $x=1$ , $E_{\max}=m_\mu/2$ ; $f(y)\to1$	the clean Michel spectrum; corrections are $O(m_e^2/m_\mu^2)\sim2\times10^{-5}$
Pure $V-A$	Michel parameters $\rho=3/4,\ \delta=3/4,\ \xi=1,\ \eta=0$	the Standard Model prediction the spectrum shape tests
Scalar/tensor admixture	$\rho\neq3/4$ ; spectrum shape distorted	a would-be signal of physics beyond $V-A$ ; none seen to date
Heavier lepton (scale up $m$ )	$\Gamma\propto m^5$ ; e.g. $\tau$ decay leptonic width	the same formula predicts the $\tau$ lifetime once branching is included

Common trapThe weak interaction is intrinsically feeble

It is not. The temptation is to read the tiny value $G_F\sim10^{-5}\,\mathrm{GeV}^{-2}$ as a small coupling, but $G_F$ is not a coupling — it is a coupling divided by a mass squared, $G_F/\sqrt2=g_W^2/8M_W^2$ . The actual coupling is $\alpha_W=g_W^2/4\pi\approx1/30$ , which is larger than the electromagnetic $\alpha=1/137$ . The entire apparent weakness of the low-energy weak interaction comes from the enormous $M_W^2\approx(80\,\mathrm{GeV})^2$ sitting in the propagator denominator: at momentum transfers well below $M_W$ , the amplitude is throttled by $1/M_W^2$ . Push the energy up to the $W$ mass — at the LHC, say — and the "weak" force is comparable in strength to electromagnetism, which is exactly the observation that motivated electroweak unification. So $G_F$ small means "heavy mediator", not "weak coupling".

Worked examples

1 · The muon lifetime from first principles (the flagship number)

Using $G_F=1.16637\times10^{-5}\,\mathrm{GeV}^{-2}$ and $m_\mu=105.658$ MeV, predict the muon lifetime and compare with the measured $2.1970\ \mu$ s.

Step 1. Assemble the fifth power of the mass in GeV: $m_\mu=0.105658$ GeV, so $m_\mu^5=1.317\times10^{-5}\,\mathrm{GeV}^5$ . Square the Fermi constant: $G_F^2=1.3604\times10^{-10}\,\mathrm{GeV}^{-4}$ .

Step 2. Form the width $\Gamma=\dfrac{G_F^2 m_\mu^5}{192\pi^3}$ . The denominator is $192\pi^3=5953.2$ . So $\Gamma=\dfrac{(1.3604\times10^{-10})(1.317\times10^{-5})}{5953.2}=\mathbf{3.009\times10^{-19}\ GeV}$ .

Step 3. Convert to a lifetime with $\tau=\hbar/\Gamma$ and $\hbar=6.5821\times10^{-25}\,\mathrm{GeV\,s}$ : $\tau_\mu=\dfrac{6.5821\times10^{-25}}{3.009\times10^{-19}}=\mathbf{2.187\ \mu s}$ .

Step 4. Compare: the measured value is $2.1970\ \mu$ s, so the tree-level formula is high by about $0.5\%$ . That residual is precisely the electron-mass phase-space factor $f\approx0.99981$ (negligible) plus the QED radiative correction $\approx0.9958$ — apply them and the prediction lands on the measurement. Tree level gives $\tau_\mu=2.19\ \mu$ s; the $0.5\%$ gap is the known radiative correction, and closing it is how $G_F$ is extracted to sub-ppm.

2 · The electron endpoint energy (kinematics)

What is the maximum energy the electron can carry away in $\mu^-\to e^-\bar\nu_e\nu_\mu$ ?

Step 1. The electron is fastest when the two neutrinos recoil together opposite to it, so the neutrino pair acts like a single massless object and the decay is effectively two-body: $\mu\to e + (\nu\nu)$ with the pair massless. Energy-momentum conservation in the muon rest frame then fixes the electron energy.

Step 2. For a two-body split of a mass $m_\mu$ into a particle of mass $m_e$ and a massless system, $E_{\max}=\dfrac{m_\mu^2+m_e^2}{2m_\mu}$ . Insert numbers: $E_{\max}=\dfrac{(105.658)^2+(0.511)^2}{2(105.658)}=\dfrac{11163.9}{211.316}=\mathbf{52.83\ MeV}$ .

Step 3. With the electron mass neglected this is just $m_\mu/2=52.83$ MeV — the electron mass shifts it by only $2.5$ keV. In the reduced variable this is $x=1$ , the sharp right edge of the Michel spectrum in the figure. The electron endpoint is $E_{\max}=m_\mu/2\approx 52.8$ MeV, and its sharpness is a clean test of the spectrum shape.

3 · Predicting the tau lifetime from the fifth-power law (lepton universality)

Assume the same $G_F$ governs the tau. Given $m_\tau=1776.86$ MeV, $\tau_\mu=2.197\ \mu$ s, and that the tau decays to $e\bar\nu_e\nu_\tau$ with branching $B_e=17.82\%$ , predict the tau lifetime.

Step 1. The leptonic partial width scales as $m^5$ : $\Gamma(\tau\to e\bar\nu_e\nu_\tau)/\Gamma(\mu\to e\bar\nu_e\nu_\mu)=(m_\tau/m_\mu)^5$ . Compute $(m_\mu/m_\tau)^5=(105.658/1776.86)^5=(0.059463)^5=\mathbf{7.43\times10^{-7}}$ .

Step 2. The tau's total width is larger than its leptonic partial width because it also decays hadronically; the electron channel is only a fraction $B_e$ . So $\tau_\tau=\tau_\mu\,(m_\mu/m_\tau)^5\,B_e$ .

Step 3. Evaluate: $\tau_\tau=(2.197\times10^{-6}\,\mathrm s)(7.43\times10^{-7})(0.1782)=\mathbf{2.91\times10^{-13}\ s}=291\ \mathrm{fs}$ .

Step 4. The measured tau lifetime is $290.3$ fs. The prediction agrees to better than $0.3\%$ , using nothing but the muon lifetime, the mass ratio to the fifth power, and the electron branching fraction. The $m^5$ law and a single universal $G_F$ predict $\tau_\tau\approx291$ fs versus $290$ fs measured — a direct confirmation of lepton universality.

Physics of the portrait

The muon-decay spectrum is where a young mathematical physicist gave experimenters a ruler for the weak interaction's chiral structure — years before anyone knew that structure was $V-A$ .

Origins · 1 minLouis Michel and how this was discovered

Louis Michel

1923–1999 · the general theory of the muon-decay electron spectrum

Louis Michel was a French mathematical physicist who, in 1949–1950, asked a sharp question: if a muon decays into an electron and two unseen light neutrals through some unknown four-fermion interaction, what does the electron energy spectrum look like for the most general possible interaction? Rather than assume a particular coupling, he wrote down every Lorentz-invariant local four-fermion term — scalar, pseudoscalar, vector, axial-vector, and tensor — and showed that the entire spectrum shape and its dependence on muon polarisation are captured by just four numbers, now universally called the Michel parameters $\rho,\ \eta,\ \xi,\ \delta$ . The isotropic energy spectrum is fixed almost entirely by $\rho$ , so measuring it discriminates among the possible couplings. This was a remarkable act of foresight: Michel handed the community a model-independent template in 1950, and when parity violation was found in 1957 and the $V-A$ theory of Feynman–Gell-Mann and Sudarshan–Marshak followed in 1958, the muon spectrum was already waiting to confirm $\rho=3/4$ . Modern measurements give $\rho=0.74977\pm0.00012$ , one of the most precise confirmations of the Standard Model's chiral structure. His name is fixed forever to the spectrum an undergraduate now derives in an afternoon.

History

1934

— Fermi proposes the four-fermion contact theory of beta decay

1937

— the muon is found in cosmic rays by Neddermeyer and Anderson

1950

— Michel derives the general electron energy spectrum and its parameters

1957

— parity violation is discovered; the V−A interaction is fixed the following year

2011

— the MuLan experiment measures the muon lifetime to 1 ppm, pinning G_F

Going further

Texts. Thomson, Modern Particle Physics, §11.5–11.6 (the full muon-decay calculation) · Griffiths, Introduction to Elementary Particles, 2nd ed., §9.1–9.2 (Eqs. 9.15–9.40) · Halzen & Martin, Quarks and Leptons, §12.6–12.8 · Peskin & Schroeder, An Introduction to Quantum Field Theory, §5.3 and Problem 5.4 (trace evaluation and phase space) · Schwartz, Quantum Field Theory and the Standard Model, Ch. 29. The key results assembled in this article are: (1) $G_F/\sqrt2=g_W^2/8M_W^2$ and $\alpha_W=g_W^2/4\pi\approx1/30$ ; (2) $\langle|\mathcal M|^2\rangle=64G_F^2(p_\mu\!\cdot p_{\bar\nu_e})(p_e\!\cdot p_{\nu_\mu})$ ; (3) the Michel spectrum $2x^2(3-2x)$ integrates to unity; (4) $\Gamma=G_F^2 m_\mu^5/192\pi^3$ gives $\tau_\mu=2.19\ \mu$ s at tree level; (5) $E_{\max}=(m_\mu^2+m_e^2)/2m_\mu=52.8$ MeV; (6) the $\tau$ lifetime $291$ fs from $m^5$ scaling and $B_e=17.82\%$ .

Prerequisites. Charged-Current Weak Interaction · The V−A Structure · The W Boson Propagator · Trace Techniques · Decay Rates and Phase Space. Leads to. The Fermi Constant · Lepton Universality · Weak Interactions of Quarks · Electroweak Unification.

Ask the tutor: Want the neutrino phase-space integral done line by line, or how the Michel parameter rules out right-handed currents?

Why does muon decay produce two neutrinos rather than one?

Because two separate conservation laws must both hold. Muon-family number requires the muon-neutrino $\nu_\mu$ to carry away the muon's identity, and electron-family number requires that creating an electron be balanced by an electron-antineutrino $\bar\nu_e$ . A single neutrino could not do both jobs at once — it would carry the wrong lepton numbers for one family or the other. The two-neutrino final state is also why the electron comes out with a continuous spectrum rather than a fixed energy: with three bodies sharing the released energy, the electron can take anything from nearly zero up to the $52.8$ MeV endpoint. A two-body decay like $\mu\to e\gamma$ , by contrast, would give a monoenergetic electron — and is forbidden precisely because it violates lepton-family number.

Is the weak coupling really stronger than electromagnetism?

Yes, at the level of the fundamental coupling constant. The weak fine-structure constant is $\alpha_W=g_W^2/4\pi\approx1/30$ , compared with the electromagnetic $\alpha\approx1/137$ . What makes the weak interaction look weak at everyday energies is not the coupling but the mediator: the $W$ boson weighs $80$ GeV, and its propagator suppresses the amplitude by $1/M_W^2$ whenever the momentum transfer is small. In muon decay the transfer is tens of MeV, so the rate suppression is of order $(m_\mu/M_W)^4\sim10^{-12}$ . Raise the energy to $M_W$ and the suppression disappears — at the LHC the "weak" and electromagnetic interactions have comparable strength, which is the whole point of electroweak unification.

What fixes the electron endpoint at 52.8 MeV?

Pure kinematics. In the muon rest frame the total available energy is $m_\mu c^2=105.7$ MeV. The electron carries the most energy when the two neutrinos both recoil in the opposite direction, so that the neutrino pair behaves like a single massless particle recoiling against the electron. That configuration is a two-body decay, and two-body kinematics fixes the electron energy at $E_{\max}=(m_\mu^2+m_e^2)/2m_\mu$ , which is $52.83$ MeV — essentially $m_\mu/2$ since the electron is so light. The sharpness of this edge is experimentally valuable: any shift would signal a massive neutrino or an unexpected interaction, so the endpoint doubles as an energy reference.

How is G_F actually measured?

By timing muons. You stop positive muons in a target, start a clock, and record when each decay positron appears; the exponential decay curve gives the lifetime $\tau_\mu$ . Inverting the rate formula (with the calculated electron-mass and radiative corrections applied) turns $\tau_\mu$ into $G_F$ . The MuLan experiment at PSI did exactly this with roughly $10^{12}$ recorded decays, reaching $\tau_\mu=2\,196\,980.3$ ps at $1$ -ppm precision, which yields $G_F=1.1663787(6)\times10^{-5}\,\mathrm{GeV}^{-2}$ . Because muon decay is theoretically clean — no hadrons, fully calculable QED corrections — $G_F$ is one of the most precisely known quantities in physics, and it is the reference input tying together the electroweak sector.

Why is the decay μ⁻ → e⁻γ not seen instead?

Because it would violate lepton-family number, which the Standard Model conserves to an extraordinary degree. The ordinary decay $\mu^-\to e^-\bar\nu_e\nu_\mu$ carefully balances the muon and electron family numbers using two neutrinos; the radiative mode $\mu^-\to e^-\gamma$ has no neutrinos and so changes muon number by one and electron number by one, which no Standard Model vertex allows. Neutrino oscillations do induce it at an utterly negligible level (branching $\sim10^{-54}$ ), so any observation at the $10^{-13}$ – $10^{-14}$ level that experiments like MEG-II can reach would be a clean signal of new physics — supersymmetry, heavy neutral leptons, or new scalars. The absence of $\mu\to e\gamma$ is therefore one of the tightest experimental constraints on physics beyond the Standard Model.

What exactly is the Michel parameter ρ, and why is it 3/4?

The Michel parameter $\rho$ is the number that controls the curvature of the isotropic electron energy spectrum. Michel showed that for the most general four-fermion interaction the spectrum has the form $x^2[3(1-x)+\tfrac{2}{3}\rho(4x-3)]$ (neglecting the electron mass), so $\rho$ dials the shape between different assumed couplings. Pure $V-A$ predicts $\rho=3/4$ exactly, which turns the bracket into the clean $2x^2(3-2x)$ we integrated. A pure scalar or pseudoscalar interaction would give a different curvature. Measuring $\rho=0.74977\pm0.00012$ is thus a direct, model-independent confirmation that the charged-current interaction is left-handed vector-minus-axial, with no room for a significant scalar or tensor admixture. It is one of the cleanest quantitative tests of the Standard Model's chiral structure.