Find the velocity of the muon in Ex. 12.8.

Write the expression for the conservation of momentum.

$$\begin{gathered} p_{before}=p_{after} \\ {p}_{before}=p_{\mu }+p_{\mu } \end{gathered}$$

As the pion is at rest, the initial momentum will be zero. Hence, the above equation becomes,

role="math" localid="1654751319282" $$\begin{gathered} 0=p_{\mu }+p_v \\ {\mathit{p}}_{\mathbf{\mu }}\mathbf{=}\mathbf{-}{\mathit{p}}_{\mathbf{v}} \end{gathered}$$

Write the expression for the conservation of energy.

role="math" localid="1654751332433" $$\begin{gathered} E_{before}=E_{after} \\ {\mathit{m}}_{\mathbf{\pi }}{\mathit{c}}^{\mathbf{2}}\mathbf{=}{\mathit{E}}_{\mathbf{\mu }}\mathbf{+}{\mathit{E}}_{\mathbf{v}} \end{gathered}$$ …… (1)

Here, it is known that:

$$\begin{gathered} E_v=\left|p_c\right|c \\ {E}_v=\left|p_{\mu }\right|c \end{gathered}$$ …… (2)

Using equation 12.54, write the value of $p_{\mu }$.

$p_{\mu }=\left[\frac{\sqrt{{E^2}_{\mu }-{m^2}_{\mu }{c}^4}}{c}\right]$

Substitute $p_{\mu }=\left[\frac{\sqrt{{E^2}_{\mu }-{m^2}_{\mu }{c}^4}}{c}\right]$for $p_{\mu }$in equation (2).

$$\begin{gathered} m_{\pi }{c}^2=E_{\mu }+\sqrt{{E^2}_{\mu }-{m^2}_{\mu }{c}^4} \\ {m}_{\pi }{c}^2={E^2}_{\mu }+\left({E^2}_{\mu }-{m^2}_{\mu }{c}^4\right) \\ {c}^2\left({m^2}_{\pi }+{m^2}_{\mu }\right)=2{E^2}_{\mu } \\ {E}_{\mu }=\frac{\left({m^2}_{\pi }+{m^2}_{\mu }\right)c^2}{2m_{\pi }} \end{gathered}$$

Write the formula for the Lorentz contraction.

$y=\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}\dots \dots \left(3\right)$

Here, the value of $y$in terms of $m_{\pi }\mathrm{and}{m}_{\mu }$is given as:

$y=\frac{{m^2}_{\pi }+{m^2}_{\mu }}{2m_{\pi }{m}_{\mu }}$

Substitute $\frac{{m^2}_{\pi }+{m^2}_{\mu }}{2m_{\pi }{m}_{\mu }}$for $y$in equation (3).

$$\begin{gathered} \frac{{m^2}_{\pi }+{m^2}_{\mu }}{2m_{\pi }{m}_{\mu }}=\frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}} \\ \frac{1}{\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}}=\frac{1}{\frac{{m^2}_{\pi }+{m^2}_{\mu }}{2m_{\pi }{m}_{\mu }}} \\ 1-\frac{v^2}{c^2}=\frac{4{m^2}_{\pi }{m^2}_{\mu }}{{\left({m^2}_{\pi }+{m^2}_{\mu }\right)}^2} \\ \frac{v^2}{c^2}=1-\frac{4{m^2}_{\pi }{m^2}_{\mu }}{{\left({m^2}_{\pi }+{m^2}_{\mu }\right)}^2} \end{gathered}$$

On further solving,

role="math" localid="1654753313768" $$\begin{gathered} \frac{v^2}{c^2}=\frac{{m^4}_{\pi }{m^4}_{\mu }+2{m^2}_{\pi }+{m^2}_{\mu }-4{m^2}_{\pi }+{m^2}_{\mu }}{{\left({m^2}_{\pi }+{m^2}_{\mu }\right)}^2} \\ \frac{v^2}{c^2}=\frac{{m^4}_{\pi }{m^4}_{\mu }-2{m^2}_{\pi }{m^2}_{\mu }}{{\left({m^2}_{\pi }+{m^2}_{\mu }\right)}^2} \\ \frac{v^2}{c^2}=\frac{{\left({m^2}_{\pi }-{m^2}_{\mu }\right)}^2}{{\left({m^2}_{\pi }+{m^2}_{\mu }\right)}^2} \\ v=\left[\frac{\left({m^2}_{\pi }-{m^2}_{\mu }\right)}{\left({m^2}_{\pi }+{m^2}_{\mu }\right)}\right]c \end{gathered}$$

Therefore, the velocity of the muon is$v=\left[\frac{\left({m^2}_{\pi }-{m^2}_{\mu }\right)}{\left({m^2}_{\pi }+{m^2}_{\mu }\right)}\right]c$.