Chapter 26, Problem 74GP

To determine

To show: The kinetic energy of the muon is ${\text{KE}}_{\mu }=\frac{{\left(m_{\pi }-m_{\mu }\right)}^2{c}^2}{2m_{\pi }}$

Explanation of Solution

Given:

Rest mass of pi-meson is $m_{\pi }$

Rest mass of muon is $m_{\mu }$

Rest mass of neutrino is $m_{\nu }=0$

Formula used:

Relativistic energy is given by:

  $E=\sqrt{p^2{c}^2+m^2{c}^4}$

Calculation:

According to the given statement, pi meson is at rest so the momentum is zero.

Using the law of conservation of energy,

  $E_{\pi }=E_{\mu }+E_{\nu }$

Now, the energy of each particle is given by:

  $\begin{array}{l}{E}_{\pi }=\sqrt{p_{\pi }^2{c}^2+m_{\pi }^2{c}^4}\\ E_{\pi }=\sqrt{0+m_{\pi }^2{c}^4}=m_{\pi }{c}^2\end{array}$

Also, according to the momentum conservation law, $p_{\nu }=p_{\mu }=p$

Hence, the energy for muon and neutrino can be given as follows:

  $\begin{array}{l}{E}_{\mu }=\sqrt{p_{\mu }^2{c}^2+m_{\mu }^2{c}^4}\\ E_{\mu }=\sqrt{p^2{c}^2+m_{\mu }^2{c}^4}\end{array}$

  $\begin{array}{l}{E}_{\nu }=\sqrt{p_{\nu }^2{c}^2+m_{\nu }^2{c}^4}\\ E_{\nu }=\sqrt{p^2{c}^2+0}=pc\end{array}$

Using all the values and substituting them into the energy conservation law,

  $\begin{array}{l}{E}_{\pi }=E_{\mu }+E_{\nu }\\ m_{\pi }{c}^2=\sqrt{p^2{c}^2+m_{\mu }^2{c}^4}+pc\\ m_{\pi }{c}^2=\sqrt{E_{\nu }^2+m_{\mu }^2{c}^4}+E_{\nu }\\ {\left(m_{\pi }{c}^2-E_{\nu }\right)}^2=\left(E_{\nu }^2+m_{\mu }^2{c}^4\right)\\ E_{\nu }=\frac{1}{2}{m}_{\pi }{c}^2-\frac{1}{2}\frac{m_{\mu }^2{c}^2}{m_{\pi }}\end{array}$

Now, the kinetic energy for muon can be calculated as follows:

  $\begin{array}{l}{\text{KE}}_{\mu }=m_{\pi }{c}^2-E_{\nu }\\ {\text{KE}}_{\mu }=m_{\pi }{c}^2-\left(\frac{1}{2}{m}_{\pi }{c}^2-\frac{1}{2}\frac{m_{\mu }^2{c}^2}{m_{\pi }}\right)\\ {\text{KE}}_{\mu }=\frac{\left(m_{\pi }^2-m_{\mu }^2\right)c^2}{2m_{\pi }}-m_{\mu }{c}^2\\ {\text{KE}}_{\mu }=\frac{{\left(m_{\pi }-m_{\mu }\right)}^2{c}^2}{2m_{\pi }}\end{array}$

Conclusion:

Thus, the kinetic energy of the muon is ${\text{KE}}_{\mu }=\frac{{\left(m_{\pi }-m_{\mu }\right)}^2{c}^2}{2m_{\pi }}$