Still Another Conservation Law — Question 1

Question 1

Associated production preserves strangeness in a strong collision

For $\pi^-+p\to K^0+\Lambda$, assign S=0 to the initial particles, $S(K^0)=+1$ and $S(\Lambda)=-1$. Baryon numbers are 0 for mesons and 1 for baryons; both final particles are neutral.

1. Check charge, baryon number and strangeness.

2. Explain why replacing the final state by $\pi^0+\Lambda$ excludes a strong or electromagnetic reaction.

Question 1 - Solution

Associated production preserves strangeness in a strong collision

(a) Method and calculation

$$\boxed{Q/e:\ -1+1=0+0,\quad B:\ 0+1=0+1}.$$$$\boxed{S_i=0,\quad S_f=+1-1=0}.$$

(b) Interpretation and further calculation

$$\pi^0+\Lambda:\quad S_f=0-1=-1\Rightarrow\boxed{\Delta S=-1}.$$$$\boxed{\Delta S\ne0\Rightarrow\text{not strong or electromagnetic}}.$$

Physical check

Strangeness is additive. Conservation permits the first channel but does not specify its cross section; weak interactions can change strangeness.