Chapter 3

2013301020086
王雨祺---

4.
(a) According to prerelativistic physics, the muon will travel $l=vt=659m$, so it cannot get to ground level.
(b) Using relativistic physics, the time measured in the lab framework is :

$$t=\gamma \tau$$
where $\tau=2.2\times 10^{-6}s, \gamma=1/\sqrt{1-v^2/c^2}, v=0.998c$, so $t=3.48\times 10^{-5}s$, So the muon will travel , so it can get ground.
(c)The pion will travel $l_{\pi}=v\gamma\tau_{\pi}=123m$, so it cannot get the ground.

16.
In the lab frame, before collision:

$$p_{before}=(\frac{E_A}{c}+m_Bc,\mathbf{p_A})$$
where $\mathbf{p_A}$ is the momentum of A in the lab frame. Then we have:
$$p^2_{before}=(\frac{E_A}{c}+m_Bc)^2-p_A^2=(\frac{E_A^2}{c^2}-p_A^2)+2E_Am_B+m_B^2c^2=2E_Am_B+(m_A^2+m_B^2)c^2$$
In the CM frame after collision:
$$p_{after}=(Mc,0,0,0)$$
where $M=\sum_{i=1}^{n}m_i$. The total momentum is conserved, so we have:
$$p_{before}^2=p_{after}^2$$
which means that $E_A=\frac{M^2-m_A^2-m_B^2}{2M_B}c^2$

17.
(a)$E=\frac{(2m_p+m_{\pi^0})^2-2m_p^2}{2m_p}c^2=1218MeV$
(b)$E=\frac{(2m_p+m_{\pi^+}+m_{\pi^-})^2-2m_p^2}{2m_p}c^2=1538MeV$
(c)$E=\frac{(2m_p+m_n)^2-m_{\pi^-}^2-m_p^2}{2m_p}c^2=3747MeV$
(d)$E=\frac{(m_{K^0}+m_{\Sigma^0})^2-m_{\pi^-}^2-m_p^2}{2m_p}c^2=1043MeV$
(e)$E=\frac{(m_p+m_{\Sigma^+}+m_{K^0})^2-2m_p^2}{2m_p}c^2=2735MeV$

18.
For the second reaction $K^-+p\to\Omega^-+K^0+K^+$, the threshold energy of the $K^-$ is :

$$E_{K^-}=\frac{(m_{\Omega ^-} +m_{K^0}+m_{K^-})^2-m_{K^-}^2-m_p^2}{2m_p}c^2=3128.3MeV$$

19.
(a)$p_A^{\nu}=p_B^{\nu}+p_C^{\nu}$, put $p_B^{\nu}$ to the other side $p_A^{\nu}-p_B^{\nu}=p_C^{\nu}$ and squre it, we have:

$$p_A^2+p_B^2-2\frac{E_A}{c}\frac{E_B}{c}+2\mathbf{p_A}\cdot\mathbf{p_B}=p_C^2$$
where $E_A=m_Ac^2, p_{\alpha}^2=m_{\alpha}^2c^2, \alpha=A,B,C, \mathbf{p_A}=0$, so we get:
$$E_B=\frac{m_A^2+m_B^2-m_C^2}{2m_A}c^2$$
in the same way we get:
$$E_C=\frac{m_A^2+m_C^2-m_B^2}{2m_A}c^2$$
(b) $$\frac{E_B^2}{c^2}=\mathbf{p_B^2}+m_B^2c^2$$
$$E_B=\frac{m_A^2+m_B^2-m_C^2}{2m_A}c^2$$
so $\mathbf{p_B^2}=\frac{m_A^4+m_B^4+m_C^4-2m_A^2m_B^2-2m_A^2m_C^2-2m_C^2m_B^2}{4m_A^2}c^2=\frac{c\lambda(m_A^2,m_B^2,m_C^2)}{2m_A}$
where $\lambda(x,y,z)=x^2+y^2+z^2-2xy-2xz-2yz$.
Since $\mathbf{p_A}=0$, we have $\mathbf{p_C^2}=\mathbf{p_B^2}$.
(c)In the case $m_A=m_B+m_C$, there is no mass transfered to kinetic energy. $m_A<m_B+m_C$ is forbidden by the energy conservation law in the CM frame.

20.
(a)$E_{\mu}=109.8MeV, E_{\nu}=29.8MeV$
(b)$E_{\gamma}=67.5MeV$
(c)$E_{\pi^+}=248.1MeV, E_{\pi^0}=245.6MeV$
(d)$E_p=943.6MeV, E_{\pi^-}=172.1MeV$
(e)$E_{\Lambda}=1135MeV, E_{K^-}=536.6MeV$

22.
(a)The minimum energy is $E_{min}=m_Bc^2$ at rest.
B will have maximum energy in the case where all particles other than B go to the opposite direction where B go as a unit, which means they behave as a single particle, $M=m_C+m_D+\cdots$.
So we have $E_{max}=\frac{m_A^2+m_B^2-M^2}{2m_A}c^2$.
(b)$E_{min}=m_ec^2=0.511MeV, E_{max}=\frac{m_{\mu}^2+m_e^2}{2m_{\mu}}c^2=52.8MeV$

25.
(a)

$$s+t+u=[p_A^2+p_B^2+p_C^2+p_D^2+2p_A\cdot(p_A+p_B-p_C-p_D)]/c^2$$
we also have $p_A+p_B=p_C+p_D$, so we get:
$$s+t+u=(p_A^2+p_B^2+p_C^2+p_D^2)/c^2=m_A^2+m_B^2+m_C^2+m_D^2$$
(b)
$$sc^2=p_A^2+p_B^2+2p_A\cdot p_B=p_A^2+p_B^2+2\frac{E_A}{c}\frac{E_B}{c}-2\mathbf{p_A}\cdot\mathbf{p_B}$$
In the CM frame $\mathbf{p_B}=-\mathbf{p_A}$, so；
$$sc^2=p_A^2+p_B^2+2p_A\cdot p_B=p_A^2+p_B^2+2\frac{E_A}{c}\frac{E_B}{c}-2\mathbf{p_A^2}$$
we also have
$$\mathbf{p_A^2}=\frac{E_A^2}{c^2}-m_A^2c^2$$
$$\frac{E_B^2}{c^2}-m_B^2c^2=\mathbf{p_B^2}=\mathbf{p_A^2}=\frac{E_A^2}{c^2}-m_A^2c^2$$
so $$\frac{E_B}{c}=\sqrt{\frac{E_A^2}{c^2}-m_A^2c^2+m_B^2c^2}$$
put all together we get :
$$E_A^{CM}=\frac{s+m_A^2-m_B^2}{2\sqrt{s}}c^2$$
(c)In the lab frame $E_B=m_Bc^2$ and $\mathbf{p_B}=0$, so we have:
$$sc^2=p_A^2+p_B^2+2\frac{E_A}{c}\frac{m_Bc^2}{c}=m_A^2c^2+m_B^2c^2+2m_BE_A$$
so $E_A^{lab}=\frac{s-m_A^2+m_B^2}{2m_B}c^2$.
(d)
$$E_{tot}=E_A^{CM}+E_B^{CM}=\frac{s+m_A^2-m_B^2}{2\sqrt{s}}c^2+\frac{s+m_B^2-m_A^2}{2\sqrt{s}}c^2=\sqrt{s}c^2$$

26.

$$s=(2p_A^2+2p_A^{(1)}\cdot p_A^{(2)})/c^2=(2m^2c^2+2\frac{E^2}{c^2}+2\mathbf{p^2})/c^2=(4m^2c^2+4\mathbf{p^2})/c^2$$
$$t=(2p_A^2+2p_A^{(1)}\cdot p_A^{(3)})/c^2=(2m^2c^2-2\frac{E^2}{c^2}+2\mathbf{p^2}cos\theta)/c^2=-2\mathbf{p^2}(1-cos\theta)/c^2$$
$$u=(2p_A^2+2p_A^{(1)}\cdot p_A^{(4)})/c^2=(2m^2c^2-2\frac{E^2}{c^2}-2\mathbf{p^2}cos\theta)/c^2=-2\mathbf{p^2}(1+cos\theta)/c^2$$

27.
Suppose a photon of wavelength $\lambda$ collides elastically with a charged particle with mass $m$, if the photon scatters at angle $\theta$ with wavelength $\lambda^{\prime}$ , the charged particle scatters at angle $\phi$ with momentum $p$.
Before the collision, the total momentum is :

$$p_{before}=(\frac{h}{\lambda}+mc,\frac{h}{\lambda},0,0)\tag 1$$
After the collision, the poton has the momentum :
$$p_{\gamma}=(\frac{h}{\lambda^{\prime}},\lambda^{\prime}cos\theta,0,0)\tag 2$$
the charged particle has the momentum :
$$p_m=(\frac{E}{c}, pcos\phi,-psin\phi,0)\tag 3$$
where $\frac{E}{c}=\sqrt{p^2+m^2c^2}$.
The momentum conservation law says that :
$$\frac{h}{\lambda}+mc=\frac{h}{\lambda^{\prime}}+\sqrt{p^2+m^2c^2}\tag 4$$
$$\frac{h}{\lambda}=\lambda^{\prime}cos\theta+pcos\phi\tag 5$$
$$0=\lambda^{\prime}cos\theta-psin\phi\tag 6$$
moving some terms :
$$\frac{h}{\lambda}+mc-\frac{h}{\lambda^{\prime}}=\sqrt{p^2+m^2c^2}\tag 7$$
$$\frac{h}{\lambda}-\lambda^{\prime}cos\theta=pcos\phi\tag 8$$
$$\frac{h}{\lambda^{\prime}}cos\theta=psin\phi\tag 9$$
square all :
$$\frac{h^2}{\lambda ^2}+\frac{h^2}{\lambda^{\prime2}}+m^2c^2+2mch(\frac{1}{\lambda}-\frac{1}{\lambda^{\prime}})-2\frac{h^2}{\lambda \lambda^{\prime}}=p^2+m^2c^2 \tag{10}$$
$$h^2(\frac{1}{\lambda^2}+\frac{1}{\lambda^{\prime2}}-\frac{2cos\theta}{\lambda \lambda^{\prime}})=p^2\tag {11}$$
We finally find that :
$$\lambda^{\prime}=\lambda+\frac{h}{mc}(1-cos\theta)$$