Calculate the threshold (minimum) momentum the pion must have in order for the process $\mathbf{\pi }\mathbf{+}\mathbf{p}\mathbf{\to }\mathbf{K}\mathbf{+}\mathbf{\sum }_{}$to occur. The proton p is initially at rest. Use localid="1654341712179" ${\mathbf{m}}_{\mathbf{\pi }}{\mathbf{c}}^{\mathbf{2}}\mathbf{=}\mathbf{150}\mathbf{}\mathbf{,}\mathbf{}{\mathbf{m}}_{\mathbf{k}}{\mathbf{c}}^{\mathbf{2}}\mathbf{=}\mathbf{500}\mathbf{}\mathbf{,}\mathbf{}{\mathbf{m}}_{\mathbf{p}}{\mathbf{c}}^{\mathbf{2}}\mathbf{=}\mathbf{900}\mathbf{}\mathbf{,}\mathbf{}{\mathbf{m}}_{\mathbf{\sum }\mathbf{}_{}}{\mathbf{c}}^{\mathbf{2}}\mathbf{=}\mathbf{1200}$(all in MeV). [Hint: To formulate the threshold condition, examine the collision in the center-of-momentum frame (Prob. 12.31). Answer: 1133 MeV/c]

Using equation 12.54, write the relationship between relativistic energy and momentum.

${\mathbf{E}}^{\mathbf{2}}\mathbf{-}{\mathbf{p}}^{\mathbf{2}}{\mathbf{c}}^{\mathbf{2}}\mathbf{=}{\mathbf{m}}^{\mathbf{2}}{\mathbf{c}}^{\mathbf{4}}$ .......(1)

Here, p is the momentum, E is the energy, m is the mass, and c is the speed of light.

As the photon is initially at rest, the initial momentum will be

$\mathrm{Initial}\mathrm{momentum}={\mathrm{p}}_{\mathrm{\pi }}$

$\mathrm{Write}\mathrm{the}\mathrm{expression}\mathrm{for}\mathrm{the}\mathrm{initial}\mathrm{energy}\mathrm{of}\mathrm{\pi }$

${\mathrm{E}}_{\mathrm{\pi }}=\sqrt{({\mathrm{m}}_{\mathrm{\pi }}^2{\mathrm{c}}^4+{\mathrm{p}}_{\mathrm{\pi }}^2{\mathrm{c}}^2)}$

Write the expression for the total initial energy.

role="math" localid="1654603068454" ${\mathrm{E}}_{\mathrm{in}}={\mathrm{m}}_{\mathrm{p}}{\mathrm{c}}^2+\sqrt{({\mathrm{m}}_{\mathrm{\pi }}^2{\mathrm{c}}^4+{\mathrm{p}}_{\mathrm{\pi }}^2{\mathrm{c}}^2)}$

Write the expression for the final energy.

${\mathrm{E}}_{\mathrm{f}}=({\mathrm{m}}_{\mathrm{K}}+{\mathrm{m}}_{\sum }){\mathrm{c}}^4$

It is given that:

$\mathrm{\pi }+\mathrm{p}\to \mathrm{K}+\sum$

Substitute${\mathrm{m}}_{\mathrm{p}}{\mathrm{c}}^2+\surd ({\mathrm{m}}_{\mathrm{\pi }}^2{\mathrm{c}}^4+{\mathrm{p}}_{\mathrm{\pi }}^2{\mathrm{c}}^2)\mathrm{for}{\mathrm{E}}_{\mathrm{in}},{\mathrm{p}}_{\mathrm{\pi }}^2\mathrm{and}({\mathrm{m}}_{\mathrm{K}}+{\mathrm{m}}_{\sum })\mathrm{for}\mathrm{m}\mathrm{in}\mathrm{equation}\left(1\right).$

$$\begin{gathered} ({\mathrm{m}}_{\mathrm{p}}{\mathrm{c}}^2+\sqrt{({\mathrm{m}}_{\mathrm{\pi }}^2{\mathrm{c}}^4+{\mathrm{p}}_{\mathrm{\pi }}^2{\mathrm{c}}^2{)}^2}-{\mathrm{p}}_{\mathrm{\pi }}^2{\mathrm{c}}^2=({\mathrm{m}}_{\mathrm{K}}+{\mathrm{m}}_{\sum }{)}^2{\mathrm{c}}^4 \\ {\mathrm{m}}_{\mathrm{p}}^2{\mathrm{c}}^4+2\frac{{\mathrm{m}}_{\mathrm{p}}{\mathrm{c}}^2}{{\mathrm{c}}^4}\sqrt{({\mathrm{m}}_{\mathrm{\pi }}^2{\mathrm{c}}^4+{\mathrm{p}}_{\mathrm{\pi }}^2{\mathrm{c}}^2)}\mathrm{c}+{\mathrm{m}}_{\mathrm{\pi }}^2{\mathrm{c}}^4+{\mathrm{p}}_{\mathrm{\pi }}^2{\mathrm{c}}^2=({\mathrm{m}}_{\mathrm{K}}+{\mathrm{m}}_{\sum }{)}^2{\mathrm{c}}^4 \\ \frac{2{\mathrm{m}}_{\mathrm{p}}}{\mathrm{c}}\sqrt{({\mathrm{m}}_{\mathrm{\pi }}^2{\mathrm{c}}^2+{\mathrm{p}}_{\mathrm{\pi }}^2)}=({\mathrm{m}}_{\mathrm{K}}+{\mathrm{m}}_{\sum }{)}^2-{\mathrm{m}}_{\mathrm{p}}^2-{\mathrm{m}}_{\mathrm{\pi }}^2 \\ \frac{4{\mathrm{m}}^2}{{\mathrm{c}}^2}{\mathrm{p}}_{\mathrm{\pi }}^2=({\mathrm{m}}_{\mathrm{K}}+{\mathrm{m}}_{\sum }{)}^4-2({\mathrm{m}}_{\mathrm{p}}^2+{\mathrm{m}}_{\mathrm{\pi }}^2\left)\right({\mathrm{m}}_{\mathrm{K}}+{\mathrm{m}}_{\sum }{)}^2+{\mathrm{m}}_{\mathrm{p}}^4+{\mathrm{m}}_{\mathrm{\pi }}^4+2{\mathrm{m}}_{\mathrm{p}}^2{\mathrm{m}}_{\mathrm{\pi }}^2 \end{gathered}$$

On further solving,

$$\begin{gathered} \frac{4{\mathrm{m}}^2}{{\mathrm{c}}^2}{\mathrm{p}}_{\mathrm{\pi }}^2=({\mathrm{m}}_{\mathrm{K}}+{\mathrm{m}}_{\sum }{)}^4-2({\mathrm{m}}_{\mathrm{p}}^2+{\mathrm{m}}_{\mathrm{\pi }}^2\left)\right({\mathrm{m}}_{\mathrm{K}}+{\mathrm{m}}_{\sum }{)}^2+{\mathrm{m}}_{\mathrm{p}}^4+{\mathrm{m}}_{\mathrm{\pi }}^4+2{\mathrm{m}}_{\mathrm{p}}^2{\mathrm{m}}_{\mathrm{\pi }}^2 \\ \frac{4{\mathrm{m}}^2}{{\mathrm{c}}^2}{\mathrm{p}}_{\mathrm{\pi }}^2=({\mathrm{m}}_{\mathrm{K}}+{\mathrm{m}}_{\sum }{)}^4-2({\mathrm{m}}_{\mathrm{p}}^2+{\mathrm{m}}_{\mathrm{\pi }}^2\left)\right({\mathrm{m}}_{\mathrm{K}}+{\mathrm{m}}_{\sum }{)}^2+({\mathrm{m}}_{\mathrm{p}}^2-{\mathrm{m}}_{\mathrm{\pi }}^2{)}^2 \\ {\mathrm{p}}_{\mathrm{\pi }}=\frac{\mathrm{c}}{2{\mathrm{m}}_{\mathrm{p}}}\sqrt{({\mathrm{m}}_{\mathrm{K}}+{\mathrm{m}}_{\sum }{)}^4-2({\mathrm{m}}_{\mathrm{p}}^2+{\mathrm{m}}_{\mathrm{\pi }}^2\left)\right({\mathrm{m}}_{\mathrm{K}}+{\mathrm{m}}_{\sum }{)}^2+({\mathrm{m}}_{\mathrm{p}}^2-{\mathrm{m}}_{\mathrm{\pi }}^2{)}^2} \\ {\mathrm{p}}_{\mathrm{\pi }}\frac{1}{\left(2{\mathrm{m}}_{{\mathrm{pc}}^2}\right)\mathrm{c}}\sqrt{\frac{\left({\mathrm{m}}_{\mathrm{K}}{\mathrm{c}}^2+{\mathrm{m}}_{\sum }{\mathrm{c}}^2\right)-2\left[({\mathrm{m}}_{\mathrm{p}}{\mathrm{c}}^2{)}^2+({\mathrm{m}}_{\mathrm{\pi }}{\mathrm{c}}^2{)}^2\right]}{{\left[{\mathrm{m}}_{\mathrm{K}}{\mathrm{c}}^2+{\mathrm{m}}_{\sum }{\mathrm{c}}^2\right]}^2+{\left[({\mathrm{m}}_{\mathrm{p}}{\mathrm{c}}^2{)}^2-({\mathrm{m}}_{\mathrm{\pi }}{\mathrm{c}}^2{)}^2\right]}^2}} \end{gathered}$$.............(2

Substitute

$$\begin{gathered} {\mathrm{P}}_{\mathrm{\pi }}=\frac{1}{2\times \left(900\right)\mathrm{c}}\sqrt{{\left(500+1200\right)}^4-2{\left(900+150\right)}^2{\left(500+1200\right)}^2+{\left(900-150\right)}^2\mathrm{MeV}} \\ {\mathrm{P}}_{\mathrm{\pi }}=\frac{1}{1800\mathrm{c}}\left(2.04\times {10}^8\right)\mathrm{MeV} \\ {\mathrm{P}}_{\mathrm{\pi }}=1133\mathrm{MeV}/\mathrm{c} \\ \mathrm{Therefore},\mathrm{the}\mathrm{threshold}\mathrm{momentum}\mathrm{is}1133\mathrm{MeV}/\mathrm{c} \end{gathered}$$