# Why is the charged Kaon lifetime much shorter than the muon lifetime?

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?

• Yes, crystal clear. Thanks. May 5, 2023 at 10:03

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.