The question is quite straightforward. A $K^+$ meson can decay weakly as follows: $$ K^+ \longrightarrow \mu^+ \nu_{\mu}, $$ with a decay time of ~12 ns, whereas the weak decay of an anti muon, $$ \mu^+ \longrightarrow e \ \bar{\nu}_{e} \ \nu_{\mu} $$ has a decay time of ~2 $\mu$s. Why is there such a difference between them?

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    $\begingroup$ Yes, crystal clear. Thanks. $\endgroup$ May 5, 2023 at 10:03

1 Answer 1


There is no mystery here.

Most books, including Okun's standard go-to "Leptons and Quarks", or Commins & Buchsbaum, give you the widths for the two respective processes, and thus their ratio, $$ \frac{\Gamma_{K\to \mu \nu} }{\Gamma_{\mu\to e2\nu}}= \frac{G_F^2 192 \pi^3 \sin^2\theta_C m_\mu^2 f_K^2 m_K (1-m_\mu^2/m_K^2)^2}{G_F^2 m_\mu^5 8 \pi} \approx 116, $$ when I plug in my best numbers. The rate in the denominator is necessary by dimensional analysis, as a 1-scale problem; while the 2-body decay on top has a Cabbibo angle suppressor, as well as the K-decay constant, $f_K\sim 158$MeV, quantifying hadronization of the strangeness-changing quark weak current, as well as a Kaon mass. (Note the phase-space factor in parenthesis is not that important, as the Kaon is relatively heavy, $ m_K\sim 494$MeV.) Most good texts work the relevant amps out for you.

  • Compare this with your $$ \frac{\tau_\mu}{\tau_K/0.64}=\frac{2\cdot 0.64 }{1.2 \cdot 10^{-2}}\sim 107, $$ where the 0.64 BR for that 2-body leptonic decay mode is multiplying the total width.

The two numbers are really not that different, given the sloppy values I have plugged in...


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