Still Another Conservation Law — Question 7

Question 7

Semileptonic kaon decay links strangeness change with flavor bookkeeping

Examine $K^0\to\pi^-+e^++\nu_e$. The kaon has S=+1 and B=L=0; the pion has S=B=L=0. The positron has electron-family number -1 and the neutrino +1.

1. Check charge, baryon number, electron-family number and strangeness change.

2. If the kaon rest energy is 498 MeV, pion rest energy 140 MeV and electron rest energy .511 MeV, find the total kinetic energy available in its rest frame for a massless neutrino.

Question 7 - Solution

Semileptonic kaon decay links strangeness change with flavor bookkeeping

(a) Method and calculation

$$\boxed{Q/e:\ 0=-1+1+0,\quad B:\ 0=0}.$$$$\boxed{L_e:\ 0=-1+1,\quad\Delta S=0-(+1)=-1}.$$

(b) Interpretation and further calculation

$$\boxed{Q_{\mathrm{energy}}=498-140-.511=357.489\,\mathrm{MeV}}.$$$$\boxed{K_\pi+K_{e^+}+E_\nu=357.489\,\mathrm{MeV}}\quad\text{with a continuous partition}.$$

Physical check

The strangeness-changing semileptonic channel is weak. Three final particles allow a continuum of individual energies.