91Ó°ÊÓ

Write the expression for the conservation of momentum.

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

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

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

Write the expression for the conservation of energy.

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

Here, it is known that:

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

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

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

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

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

Write the formula for the Lorentz contraction.

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

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

$y=\frac{{m^2}_{\mathrm{π}}+{m^2}_{\mathrm{μ}}}{2m_{\mathrm{π}}{m}_{\mathrm{μ}}}$

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

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

On further solving,

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

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