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research:memos:kinematics:non-relativistic_kinematics [2017/09/29 14:32] – kobayashresearch:memos:kinematics:non-relativistic_kinematics [2020/07/29 17:10] (現在) – [Formulae for Non-relativistic Kinematics] kobayash
行 8: 行 8:
 {{:research:memos:kinematics:lab_cm_systems.gif|}} {{:research:memos:kinematics:lab_cm_systems.gif|}}
  
-Adopt units where $c=1$. In the Laboratory System (Center of Mass System), mass, momentum, total energy, and velocity of the $i$-th particle are $m_i$, $\boldsymbol{p}_i$, and $E_i$, and $\boldsymbol{v}_i$ ($m_i^*$, $\boldsymbol{p}_i^*$, $E_i^*$, and $\boldsymbol{v}_i^*$), respectively.+Adopt units where $c=1$. In the Laboratory System (Center of Mass System), mass, momentum, kinetic energy, and velocity of the $i$-th particle are $m_i$, $\boldsymbol{p}_i$, and $T_i$, and $\boldsymbol{v}_i$ ($m_i^*$, $\boldsymbol{p}_i^*$, $T_i^*$, and $\boldsymbol{v}_i^*$), respectively.
  
 In the following, the quantities $\delta_{ij}$ are defined by In the following, the quantities $\delta_{ij}$ are defined by
行 14: 行 14:
 \delta_{ij} = |\boldsymbol{v}_i|/|\boldsymbol{v}_j|, \delta_{ij} = |\boldsymbol{v}_i|/|\boldsymbol{v}_j|,
 \end{align} \end{align}
-where the subscripts refer to the particles.+where the subscripts refer to the particles. In the case of elastic scattering, $|\boldsymbol{p}_3^{*}| = |\boldsymbol{p}_1^{*}|$ as below. In addition, $\boldsymbol{p}_1^* = m_1\boldsymbol{v}_1^* = \frac{m_2}{m_1+m_2}\boldsymbol{p}_1$ and $\boldsymbol{v}_2^* = \boldsymbol{v}_{\rm cm} = \frac{\boldsymbol{p}_1} {m_1+m_2}$. From these formula, 
 + 
 +\begin{align} 
 +\delta_{23}^* = \delta_{21}^* = |\boldsymbol{v}_2^*|/|\boldsymbol{v}_1^*| = m_1/m_2. 
 +\end{align}
  
 |** Quantity **|** General Formula **|**Elastic Scattering**|**N-N Scattering (equal mass)**| |** Quantity **|** General Formula **|**Elastic Scattering**|**N-N Scattering (equal mass)**|
行 20: 行 24:
 W W
 &= m_1 + m_2\\ &= m_1 + m_2\\
-&= m_3 + m_4+&= m_3 + m_4?
 \end{align}| Same as the General formula |\begin{align} \end{align}| Same as the General formula |\begin{align}
-W= \sqrt{2(m_\mathrm{N}^2+m_\mathrm{N}E_1)}+W= 2m_\mathrm{N}?
 \end{align}| \end{align}|
 | 2. c.m. momentum before the interaction |\begin{align} | 2. c.m. momentum before the interaction |\begin{align}
-|\boldsymbol{p}_1^{*}| = \frac{1}{2W}\sqrt{\left[W^2-\left(m_1+m_2\right)^2\right]\left[W^2-\left(m_1-m_2\right)^2\right]}+\boldsymbol{p}_1^* 
 +&= \frac{m_2}{m_1+m_2}\boldsymbol{p}_1\\ 
 +&= \frac{m_1m_2}{m_1+m_2}\boldsymbol{v}_1
 \end{align}| Same as the General formula |\begin{align} \end{align}| Same as the General formula |\begin{align}
-|\boldsymbol{p}_1'| = \frac{1}{2}\sqrt{W^2-4m_\mathrm{N}^2}+\boldsymbol{p}_1' = \frac{\boldsymbol{p}_1}{2}
 \end{align}| \end{align}|
 | 3. c.m. momentum after the interaction |\begin{align} | 3. c.m. momentum after the interaction |\begin{align}
-|\boldsymbol{p}_3^{*}| = \frac{1}{2W}\sqrt{\left[W^2-\left(m_3+m_4\right)^2\right]\left[W^2-\left(m_3-m_4\right)^2\right]}+\boldsymbol{p}_3^* 
 +&= \frac{m_4}{m_3+m_4}\boldsymbol{p}_3\\ 
 +&= \frac{m_3m_4}{m_3+m_4}\boldsymbol{v}_3
 \end{align}|\begin{align} \end{align}|\begin{align}
 |\boldsymbol{p}_3^{*}| = |\boldsymbol{p}_1^{*}| |\boldsymbol{p}_3^{*}| = |\boldsymbol{p}_1^{*}|
 \end{align}|\begin{align} \end{align}|\begin{align}
-|\boldsymbol{p}_3'| = |\boldsymbol{p}_1'| = \frac{1}{2}\sqrt{W^2-4m_\mathrm{N}^2}+|\boldsymbol{p}_3'| = |\boldsymbol{p}_1'| = \frac{\boldsymbol{p}_1}{2}
 \end{align}|  \end{align}| 
 | 4. Velocity of the c.m. |\begin{align} | 4. Velocity of the c.m. |\begin{align}
-\boldsymbol{\beta}_2^* = \boldsymbol{\beta}_{\rm cm} = \frac{\boldsymbol{p}_1}{E_1+m_2}+\boldsymbol{v}_2^* = \boldsymbol{v}_{\rm cm} = \frac{\boldsymbol{p}_1}{m_1+m_2}
 \end{align}| Same as the General formula | Same as the General formula | \end{align}| Same as the General formula | Same as the General formula |
 | 5. $\gamma$ of the c.m. |\begin{align} | 5. $\gamma$ of the c.m. |\begin{align}
-\gamma_2^* = \gamma_\mathrm{cm} = \frac{E_1+m_2}{W}+\gamma_2^* = \gamma_\mathrm{cm}^* \approx 1?
 \end{align}| Same as the General formula | Same as the General formula | \end{align}| Same as the General formula | Same as the General formula |
 | 6. Maximum lab scattering angle |\begin{align} | 6. Maximum lab scattering angle |\begin{align}
-\tan\theta_{3\mathrm{max}} = \frac{1}{\gamma_2^*\sqrt{\delta_{23}^{*2}-1}}\\+\tan\theta_{3\mathrm{max}} = \frac{1}{\sqrt{\delta_{23}^{*2}-1}}\\
 \mathrm{For\ \ } \delta_{23}^* \ge 1\\ \mathrm{For\ \ } \delta_{23}^* \ge 1\\
 \mathrm{otherwise\ } \theta_{3\mathrm{max}} = 180^\circ \mathrm{otherwise\ } \theta_{3\mathrm{max}} = 180^\circ
行 50: 行 58:
 \end{align}| \end{align}|
 | 7. c.m. to lab angle ($\theta_{\rm cm} \rightarrow \theta_{\rm lab}$) |\begin{align} | 7. c.m. to lab angle ($\theta_{\rm cm} \rightarrow \theta_{\rm lab}$) |\begin{align}
-\cos\theta_3 = \frac{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}{\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}+\cos\theta_3 = \frac{\delta_{23}^*+\cos\theta_3^*}{\sqrt{\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}
 \end{align}|\begin{align} \end{align}|\begin{align}
-\tan\theta_3 &= \frac{\sin\theta_3^*}{\gamma_2^*\left(\delta_{21}^*+\cos\theta_3^*\right)}\\ +\tan\theta_3 &= \frac{\sin\theta_3^*}{\delta_{21}^*+\cos\theta_3^*}\\ 
-\tan\theta_4 &= \frac{1}{\gamma_2^*}\cot\frac{\theta_3^*}{2}+\tan\theta_4 &= \cot\frac{\theta_3^*}{2}
 \end{align}|\begin{align} \end{align}|\begin{align}
-\tan\theta_3 = \frac{\sin\theta_3^*}{\gamma_2^*\left(1+\cos\theta_3^*\right)}\\+\tan\theta_3 = \frac{\sin\theta_3^*}{1+\cos\theta_3^*}\\
 \end{align}| \end{align}|
-| 8. lab to c.m. angle transformation ($\theta_\mathrm{cm} \rightarrow \theta_\mathrm{lab}$) |\begin{align} +| 8. lab to c.m. angle transformation ($\theta_\mathrm{lab} \rightarrow \theta_\mathrm{cm}$) |\begin{align} 
-\cos\theta_3^*=\frac{\delta_{23}^*(\gamma_2^*\tan\theta_3)^2}{(\gamma_2^*\tan\theta_3)^2+1}\pm\sqrt{\left(\frac{\delta_{23}^*(\gamma_2^*\tan\theta_3)^2}{(\gamma_2^*\tan\theta_3)^2+1}\right)^2-\frac{\delta_{23}^{*2}(\gamma_2^*\tan\theta_3)^2-1}{(\gamma_2^*\tan\theta_3)^2+1}}+\cos\theta_3^*=-\frac{\delta_{23}^*\tan^2\theta_3}{\tan^2\theta_3+1}\pm\sqrt{\left(\frac{\delta_{23}^*\tan^2\theta_3}{\tan^2\theta_3+1}\right)^2-\frac{\delta_{23}^{*2}\tan^2\theta_3-1}{\tan^2\theta_3+1}}
 \end{align} Another solution (not written in the document) \begin{align} \end{align} Another solution (not written in the document) \begin{align}
-\tan\theta_{\rm cm} = \frac{\sin\theta_{\rm lab}}{\gamma_{\rm cm}\left(\cos\theta_{\rm lab}-\beta_{\rm cm}/|\boldsymbol{\beta}_3|\right)}+\tan\theta_{\rm cm} = \frac{\sin\theta_{\rm lab}}{\left(\cos\theta_{\rm lab}-v_{\rm cm}/|\boldsymbol{v}_3|\right)}
 \end{align}||| \end{align}|||
 | 9. Solid angle transformation (Jacobian) |\begin{align} | 9. Solid angle transformation (Jacobian) |\begin{align}
-\frac{d\Omega_3}{d\Omega_3^*} &= \frac{\gamma_2^*(1+\delta_{23}^*\cos\theta_3^*)}{\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]^{3/2}}\\ +\frac{d\Omega_3}{d\Omega_3^*} &= \frac{1+\delta_{23}^*\cos\theta_3^*}{\left[\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]^{3/2}}\\ 
-\frac{d\Omega_3}{d\Omega_3^*} &= \frac{\sin^3\theta_3}{\sin^3\theta_3^*}\gamma_2^*(1+\delta_{23}^*\cos\theta_3^*)+\frac{d\Omega_3}{d\Omega_3^*} &= \frac{\sin^3\theta_3}{\sin^3\theta_3^*}(1+\delta_{23}^*\cos\theta_3^*)
 \end{align}||\begin{align} \end{align}||\begin{align}
-\frac{d\Omega_3}{d\Omega_3^*} = \frac{\gamma_2^*(1+\cos\theta_3^*)}{\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(1+\cos\theta_3^*\right)^2\right]^{3/2}}+\frac{d\Omega_3}{d\Omega_3^*} = \frac{1}{2^{3/2}\sqrt{1+\cos\theta_3^*}}
 \end{align}| \end{align}|
-| 10. Relations between the $\gamma$ factors \\ N.B. $k_{12}=m_1/m_2$|\begin{align} +| 10. Relations between the $\gamma$ factors \\ N.B. $k_{12}=m_1/m_2$|Undefined?|Undefined?|Undefined?| 
-&&(\gamma_1^{*2}-1) = k_{21}^2(\gamma_2^{*2}-1)\\ +| 11. Lab quantity relations |\begin{align} 
-&&\gamma_1^* = \frac{k_{12}+\gamma_1}{\sqrt{1+k_{12}^2+2\gamma_1k_{12}}}\\ +2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3 &= \boldsymbol{p}_1^2 + \boldsymbol{p}_3^2 - \boldsymbol{p}_4^2\\ 
-&&\gamma_2^* = \frac{k_{21}+\gamma_1}{\sqrt{1+k_{21}^2+2\gamma_1k_{21}}}=\gamma_\mathrm{cm}+2|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4) &= \boldsymbol{p}_1^2 - \boldsymbol{p}_3^2 - \boldsymbol{p}_4^2
 \end{align}||\begin{align} \end{align}||\begin{align}
-\gamma_1^* = \gamma_2^*\\ +|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4) &= 0\\ 
-\gamma_1^* = \sqrt{\frac{1+\gamma_1}{2}}\\ +\theta_3+\theta_4 &= 90^\circ
-\end{align}| +
-| 11. Lab quantity relations |\begin{align} +
-2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3 = m_4^2-m_1^2-m_2^2-m_3^2+2(E_1+m_2)E_3-2E_1m_2\\ +
-2|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4) = m_3^2+m_4^2-m_1^2-m_2^2-2E_1m_2+2E_3E_4\\ +
-\end{align} The sign between $m_1$ and $m_2$ of the second formula is missing in the document.||\begin{align} +
-|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4) = T_3T_4+
 \end{align}| \end{align}|
 | 12. Maximum K.E. transfer to a stationary particle |\begin{align} | 12. Maximum K.E. transfer to a stationary particle |\begin{align}
 +T_\mathrm{max}=\frac{\left[\sqrt{m_1m_4}\pm\sqrt{m_3(m_3+m_4-m_1)}\right]^2}{(m_3+m_4)^2}T_1
 +\end{align}|\begin{align}
 T_\mathrm{max} T_\mathrm{max}
-&= \frac{2\boldsymbol{p}_1^2}{m_1(k_{12}+k_{21}+2\gamma_1)}\\ +& =\frac{4m_1m_2}{(m_1+m_2)^2}T_1\\ 
-&\approx 2m_2\boldsymbol{\beta}_1^2\gamma_1^2 +&= 2m_2 \boldsymbol{v}_\mathrm{cm}^2\\ 
-\end{align} The formula in the document is wrong?|| Empty |+&\approx 2m_2\boldsymbol{v}_1^2 
 +\end{align}|\begin{align} 
 +T_\mathrm{max}=T_1 
 +\end{align}|
  
 === Derivation of Quantity 1. Total c.m. energy === === Derivation of Quantity 1. Total c.m. energy ===
 ** The General Formula ** ** The General Formula **
  
-They are definitions.+They are definitions?
  
 ** The formula for N-N Scattering ** ** The formula for N-N Scattering **
行 98: 行 105:
 \begin{align} \begin{align}
 W W
-&= \sqrt{m_1^2+m_2^2+2m_2E_1}\\ +&= m_1 + m_2\\ 
-&= \sqrt{m_\mathrm{N}^2+m_\mathrm{N}^2+2m_\mathrm{N}E_1}\\ +&= 2m_\mathrm{N}
-&= \sqrt{2(m_\mathrm{N}^2+m_\mathrm{N}E_1)}+
 \end{align} \end{align}
  
行 106: 行 112:
 ** The General Formula ** ** The General Formula **
  
-In the center of mass frame, $\boldsymbol{p}_1^*+\boldsymbol{p}_2^*=0$. Therefore,+For the center of mass system,
 \begin{align} \begin{align}
-W +\boldsymbol{p}_1^* = \boldsymbol{p}_1 - m_1\boldsymbol{v}_\mathrm{cm}.
- &= \sqrt{(E_1^*+E_2^*)^2-(\boldsymbol{p}_1^*+\boldsymbol{p}_2^*)^2}\\ +
- &= E_1^*+E_2^*.+
 \end{align} \end{align}
-By using the following equations+From Quantity 4, the velocity of the c.m. is
 \begin{align} \begin{align}
-|\boldsymbol{p}_1| &= |\boldsymbol{p}_2|,\\ +\boldsymbol{v}_{\rm cm} = \frac{\boldsymbol{p}_1}{m_1+m_2}.
-E_1^{*2} &= \boldsymbol{p}_1^{*2}+m_1^2,\\ +
-E_2^{*2} &= \boldsymbol{p}_2^{*2}+m_2^2=\boldsymbol{p}_1^{*2}+m_2^2,+
 \end{align} \end{align}
-$W$ can be written as+By substituting this formula into the formula of $\boldsymbol{p}_1^*$,
 \begin{align} \begin{align}
-W +\boldsymbol{p}_1^* 
- &= E_1^*+E_2^*\\ +&= \boldsymbol{p}_1 - m_1\frac{\boldsymbol{p}_1}{m_1+m_2}\\ 
- &= \sqrt{\boldsymbol{p}_1^{*2}+m_1^2}+\sqrt{\boldsymbol{p}_1^{*2}+m_2^2} +&= \frac{m_2}{m_1+m_2}\boldsymbol{p}_1\\ 
-\end{align} +&= \frac{m_1m_2}{m_1+m_2}\boldsymbol{v}_1
-By making square for both sides, +
-\begin{align} +
-W^2 +
- &= \boldsymbol{p}_1^{*2}+m_1^2+\boldsymbol{p}_1^{*2}+m_2^2+2\sqrt{\left(\boldsymbol{p}_1^{*2}+m_1^2\right)\left(\boldsymbol{p}_1^{*2}+m_2^2\right)}\\ +
- &= 2\boldsymbol{p}_1^{*2}+m_1^2+m_2^2+2\sqrt{\boldsymbol{p}_1^{*4}+\boldsymbol{p}_1^{*2}\left(m_1^2+m_2^2\right)+m_1^2m_2^2}\\ +
-W^2-2\boldsymbol{p}_1^{*2}-\left(m_1^2+m_2^2\right) &= 2\sqrt{\boldsymbol{p}_1^{*4}+\boldsymbol{p}_1^{*2}\left(m_1^2+m_2^2\right)+m_1^2m_2^2}.\\ +
-\end{align} +
-Again by making square for both sides, +
-\begin{align} +
-\left[W^2-2\boldsymbol{p}_1^{*2}-\left(m_1^2+m_2^2\right)\right]^2 &= 4\left[\boldsymbol{p}_1^{*4}+\boldsymbol{p}_1^{*2}\left(m_1^2+m_2^2\right)+m_1^2m_2^2\right]\\ +
-W^4+4\boldsymbol{p}_1^{*4}+\left(m_1^2+m_2^2\right)^2-4W^2\boldsymbol{p}_1^{*2}-2W^2\left(m_1^2+m_2^2\right)+4\boldsymbol{p}_1^{*2}\left(m_1^2+m_2^2\right)&= 4\boldsymbol{p}_1^{*4}+4\boldsymbol{p}_1^{*2}\left(m_1^2+m_2^2\right)+4m_1^2m_2^2\\ +
-W^4+\left(m_1^2+m_2^2\right)^2-4W^2\boldsymbol{p}_1^{*2}-2W^2\left(m_1^2+m_2^2\right)&= 4m_1^2m_2^2\\ +
-4W^2\boldsymbol{p}_1^{*2} +
-&= W^4-2W^2\left(m_1^2+m_2^2\right)+\left(m_1^2+m_2^2\right)^2-4m_1^2m_2^2\\ +
-&= W^4-2W^2\left(m_1^2+m_2^2\right)+\left(m_1^2-m_2^2\right)^2\\ +
-&= W^4-W^2\left[\left(m_1+m_2\right)^2+\left(m_1-m_2\right)^2\right]+\left(m_1^2-m_2^2\right)^2\\ +
-&= \left[W^2-\left(m_1+m_2\right)^2\right]\left[W^2-\left(m_1-m_2\right)^2\right]\\ +
-\boldsymbol{p}_1^{*2} +
-&= \frac{1}{4W^2}\left[W^2-\left(m_1+m_2\right)^2\right]\left[W^2-\left(m_1-m_2\right)^2\right]\\ +
-|\boldsymbol{p}_1^{*}| +
-&= \frac{1}{2W}\sqrt{\left[W^2-\left(m_1+m_2\right)^2\right]\left[W^2-\left(m_1-m_2\right)^2\right]}\\+
 \end{align} \end{align}
  
行 152: 行 133:
  
 \begin{align} \begin{align}
-|\boldsymbol{p}_1'| =  +\boldsymbol{p}_1' =  
-&= \frac{1}{2W}\sqrt{\left[W^2-\left(m_1+m_2\right)^2\right]\left[W^2-\left(m_1-m_2\right)^2\right]}\\ +&= \frac{m_2}{m_1+m_2}\boldsymbol{p}_1\\ 
-&= \frac{1}{2W}\sqrt{\left[W^2-\left(m_\mathrm{N}+m_\mathrm{N}\right)^2\right]\left[W^2-\left(m_\mathrm{N}-m_\mathrm{N}\right)^2\right]}\\ +&= \frac{m_\mathrm{N}}{2m_\mathrm{N}}\boldsymbol{p}_1\\ 
-&= \frac{1}{2W}\sqrt{\left[W^2-(2m_\mathrm{N})^2\right]W^2}\\ +&= \frac{\boldsymbol{p}_1}{2}
-&= \frac{1}{2}\sqrt{W^2-4m_\mathrm{N}^2}\\+
 \end{align} \end{align}
  
行 168: 行 148:
 For the elastic scattering, $m_1 = m_3$ and $m_2 = m_4$. Therefore, For the elastic scattering, $m_1 = m_3$ and $m_2 = m_4$. Therefore,
 \begin{align} \begin{align}
-|\boldsymbol{p}_3^{*}| +|\boldsymbol{p}_3^{*}| = |\boldsymbol{p}_1^{*}|
-&= \frac{1}{2W}\sqrt{\left[W^2-\left(m_3+m_4\right)^2\right]\left[W^2-\left(m_3-m_4\right)^2\right]}\\ +
-&= \frac{1}{2W}\sqrt{\left[W^2-\left(m_1+m_2\right)^2\right]\left[W^2-\left(m_1-m_2\right)^2\right]}\\ +
-&= |\boldsymbol{p}_1^{*}|+
 \end{align} \end{align}
  
行 179: 行 156:
  
 \begin{align} \begin{align}
-|\boldsymbol{p}_3'| = |\boldsymbol{p}_1'| = \frac{1}{2}\sqrt{W^2-4m_\mathrm{N}^2}+|\boldsymbol{p}_3'| = |\boldsymbol{p}_1'| = \frac{\boldsymbol{p}_1}{2}
 \end{align} \end{align}
  
行 185: 行 162:
 ** The General Formula ** ** The General Formula **
  
-From the notes below, the velocity of the c.m. $\boldsymbol{\beta}_{\rm cm}$ of two moving particles is written as+From the notes below, the velocity of the c.m. $\boldsymbol{v}_{\rm cm}$ of two moving particles is written as
  
 \begin{align} \begin{align}
-\boldsymbol{\beta}_{\rm cm} = \frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{E_1+E_2}.+\boldsymbol{v}_{\rm cm} = \frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{m_1+m_2}.
 \end{align} \end{align}
  
 If the second particle is not moving, $\boldsymbol{p}_2=0$ and $E_2=m_2$. Therefore, If the second particle is not moving, $\boldsymbol{p}_2=0$ and $E_2=m_2$. Therefore,
 \begin{align} \begin{align}
-\boldsymbol{\beta}_{\rm cm} = \frac{\boldsymbol{p}_1}{E_1+m_2}.+\boldsymbol{v}_{\rm cm} = \frac{\boldsymbol{p}_1}{m_1+m_2}.
 \end{align} \end{align}
  
-In the center of mass system, $\boldsymbol{\beta}_\mathrm{cm}=\boldsymbol{\beta}_2^*$. As a result,+In the center of mass system, $\boldsymbol{v}_\mathrm{cm}=\boldsymbol{v}_2^*$. As a result,
  
 \begin{align} \begin{align}
-\boldsymbol{\beta}_2^* = \boldsymbol{\beta}_{\rm cm} = \frac{\boldsymbol{p}_1}{E_1+m_2}+\boldsymbol{v}_2^* = \boldsymbol{v}_{\rm cm} = \frac{\boldsymbol{p}_1}{m_1+m_2}
 \end{align} \end{align}
  
行 205: 行 182:
 ** The General Formula ** ** The General Formula **
  
-In general,+In the non-relativistic limit,
 \begin{align} \begin{align}
-\gamma = \frac{1}{\sqrt{1-|\boldsymbol{\beta}|^2}}.+\gamma 
 +&= \frac{1}{\sqrt{1-|\boldsymbol{\beta}|^2}}\\ 
 +&\approx 1?
 \end{align} \end{align}
- 
-Therefore, 
- 
-\begin{align} 
-\gamma_2^* 
-&= \frac{1}{\sqrt{1-|\boldsymbol{\beta}_2^*|^2}}\\ 
-&= \frac{1}{\sqrt{1-\left|\frac{\boldsymbol{p}_1}{E_1+m_2}\right|^2}}\\ 
-&= \frac{E_1+m_2}{\sqrt{(E_1+m_2)^2-\boldsymbol{p}_1^2}}\\ 
-&= \frac{E_1+m_2}{W}. 
-\end{align} 
- 
-In the center of the mass system, $\boldsymbol{\beta}_2^*=\boldsymbol{\beta}_\mathrm{cm}$ and $\gamma_2^* = \gamma_\mathrm{cm}$. As a result, 
- 
-\begin{align} 
-\gamma_2^* = \gamma_\mathrm{cm} = \frac{E_1+m_2}{W}. 
-\end{align} 
- 
  
 === Derivation of Quantity 6. Maximum lab scattering angle === === Derivation of Quantity 6. Maximum lab scattering angle ===
行 233: 行 195:
  
 \begin{align} \begin{align}
-\tan\theta_3 &= \frac{\sin\theta_3^*}{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}.\\+\tan\theta_3 &= \frac{\sin\theta_3^*}{\delta_{23}^*+\cos\theta_3^*}.\\
 \end{align} \end{align}
  
行 240: 行 202:
 \begin{align} \begin{align}
 \frac{d(\tan\theta_3)}{d\theta_3^*} \frac{d(\tan\theta_3)}{d\theta_3^*}
-&= \frac{\cos\theta_3^*[\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)]-\sin\theta_3^*\gamma_2^*(-\sin\theta_3^*)}{\left[\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)\right]^2}\\ +&= \frac{\cos\theta_3^*[\delta_{23}^*+\cos\theta_3^*]-\sin\theta_3^*(-\sin\theta_3^*)}{\left(\delta_{23}^*+\cos\theta_3^*\right)^2}\\ 
-&= \frac{\gamma_2^*[\delta_{23}^*\cos^2\theta_3^*+\cos^2\theta_3^*+\cos^2\theta_3^*]}{\left[\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)\right]^2}\\ +&= \frac{\delta_{23}^*\cos^2\theta_3^*+\cos^2\theta_3^*+\cos^2\theta_3^*}{\left(\delta_{23}^*+\cos\theta_3^*\right)^2}\\ 
-&= \frac{\gamma_2^*[\delta_{23}^*\cos\theta_3^*+1]}{\left[\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)\right]^2}.\\+&= \frac{\delta_{23}^*\cos\theta_3^*+1}{\left(\delta_{23}^*+\cos\theta_3^*\right)^2}.\\
 \end{align} \end{align}
  
行 256: 行 218:
 \begin{align} \begin{align}
 \tan\theta_3 \tan\theta_3
-&= \frac{\sqrt{1-(1/\delta_{23}^*)^2}}{\gamma_2^*\left(\delta_{23}^*-1/\delta_{23}^*\right)}\\ +&= \frac{\sqrt{1-(1/\delta_{23}^*)^2}}{\delta_{23}^*-1/\delta_{23}^*}\\ 
-&= \frac{\sqrt{\delta_{23}^{*2}-1}}{\gamma_2^*\left(\delta_{23}^{*2}-1\right)}\\ +&= \frac{\sqrt{\delta_{23}^{*2}-1}}{\delta_{23}^{*2}-1}\\ 
-&= \frac{1}{\gamma_2^*\sqrt{\delta_{23}^{*2}-1}}.\\+&= \frac{1}{\sqrt{\delta_{23}^{*2}-1}}.\\
 \end{align} \end{align}
  
行 267: 行 229:
 \begin{align} \begin{align}
 |\boldsymbol{p}_3'| = |\boldsymbol{p}_1'| = |\boldsymbol{p}_2'|\\ |\boldsymbol{p}_3'| = |\boldsymbol{p}_1'| = |\boldsymbol{p}_2'|\\
-|\boldsymbol{\beta}_3'| = |\boldsymbol{\beta}_1'| = |\boldsymbol{\beta}_2'|\\+|\boldsymbol{v}_3'| = |\boldsymbol{v}_1'| = |\boldsymbol{v}_2'|\\
 \end{align} \end{align}
 Therefore, Therefore,
行 278: 行 240:
 \begin{align} \begin{align}
 \tan\theta_3 \tan\theta_3
-&= \frac{\sin\theta_3^*}{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}.\\ +&= \frac{\sin\theta_3^*}{\delta_{23}^*+\cos\theta_3^*}.\\ 
-&= \frac{\sin\theta_3^*}{\gamma_2^*\left(1+\cos\theta_3^*\right)}.+&= \frac{\sin\theta_3^*}{1+\cos\theta_3^*}.
 \end{align} \end{align}
  
-Therefore, $\tan\theta_3$ becomes infinite at $\cos\theta_3^*=-1$. In that case, $\mathrm{3max}=90^\circ$.+Therefore, $\tan\theta_3$ becomes infinite at $\cos\theta_3^*=-1$. In that case, $\theta_\mathrm{3max}=90^\circ$.
  
 === Derivation of Quantity 7. c.m. to lab angle ($\theta_{\rm cm} \rightarrow \theta_{\rm lab}$) === === Derivation of Quantity 7. c.m. to lab angle ($\theta_{\rm cm} \rightarrow \theta_{\rm lab}$) ===
行 288: 行 250:
  
 \begin{align} \begin{align}
-\begin{pmatrix} +|\boldsymbol{v}_3|\cos\theta_{\rm lab} &= |\boldsymbol{v}_3^*|\cos\theta_{\rm cm}+v_{\rm cm}\\ 
-E_3\\ +|\boldsymbol{v}_3|\sin\theta_{\rm lab} &= |\boldsymbol{v}_3^*|\sin\theta_{\rm cm}
-|\boldsymbol{p}_3|\cos\theta_{\rm lab} +
-\end{pmatrix} +
-&= +
-\begin{pmatrix} +
-\gamma_{\rm cm} & \beta_{\rm cm}\gamma_{\rm cm}\\ +
-\beta_{\rm cm}\gamma_{\rm cm} & \gamma_{\rm cm} +
-\end{pmatrix} +
-\begin{pmatrix} +
-E_3^*\\ +
-|\boldsymbol{p}_3^*|\cos\theta_{\rm cm} +
-\end{pmatrix}\\ +
-|\boldsymbol{p}_3|\cos\theta_{\rm lab} &= \gamma_{\rm cm}\left(|\boldsymbol{p}_3^*|\cos\theta_{\rm cm}+\beta_{\rm cm}E_3^*\right)\\ +
-|\boldsymbol{p}_3|\sin\theta_{\rm lab} &= |\boldsymbol{p}_3^*|\sin\theta_{\rm cm}+
 \end{align} \end{align}
  
行 308: 行 257:
  
 \begin{align} \begin{align}
-\frac{|\boldsymbol{p}_3|\sin\theta_{\rm lab}}{|\boldsymbol{p}_3|\cos\theta_{\rm lab}} &= \frac{|\boldsymbol{p}_3^*|\sin\theta_{\rm cm}}{\gamma_{\rm cm}\left(|\boldsymbol{p}_3^*|\cos\theta_{\rm cm}+\beta_{\rm cm}E_3^*\right)}\\ +\frac{|\boldsymbol{v}_3|\sin\theta_{\rm lab}}{|\boldsymbol{v}_3|\cos\theta_{\rm lab}} &= \frac{|\boldsymbol{v}_3^*|\sin\theta_{\rm cm}}{|\boldsymbol{v}_3^*|\cos\theta_{\rm cm}+v_{\rm cm}}\\ 
-\tan\theta_{\rm lab} &= \frac{\sin\theta_{\rm cm}}{\gamma_{\rm cm}\left(\cos\theta_{\rm cm}+\beta_{\rm cm}(E_3^*/|\boldsymbol{p}_3^*|)\right)}\\ +\tan\theta_{\rm lab} &= \frac{\sin\theta_{\rm cm}}{\cos\theta_{\rm cm}+v_{\rm cm}/|\boldsymbol{v}_3^*|}
-\tan\theta_{\rm lab} &= \frac{\sin\theta_{\rm cm}}{\gamma_{\rm cm}\left(\cos\theta_{\rm cm}+\beta_{\rm cm}/|\boldsymbol{\beta}_3^*|\right)}\\+
 \end{align} \end{align}
-In the center of mass system, $\beta_\mathrm{cm}=|\boldsymbol{\beta}_2^*|$. Therefore, +In the center of mass system, $v_\mathrm{cm}=|\boldsymbol{v}_2^*|$. Therefore, 
 \begin{align} \begin{align}
-\beta_{\rm cm}/|\boldsymbol{\beta}_3^*| +v_{\rm cm}/|\boldsymbol{v}_3^*| 
-&= |\boldsymbol{\beta}_2^*|/|\boldsymbol{\beta}_3^*|\\+&= |\boldsymbol{v}_2^*|/|\boldsymbol{v}_3^*|\\
 &= \delta_{23}^*. &= \delta_{23}^*.
 \end{align} \end{align}
-By using this formula, $\gamma_\mathrm{cm}=\gamma_2^*$, $\theta_\mathrm{cm}=\theta_3^*$, and $\theta_\mathrm{lab}=\theta_3$,+By using this formula, $\theta_\mathrm{cm}=\theta_3^*$, and $\theta_\mathrm{lab}=\theta_3$,
 \begin{align} \begin{align}
-\tan\theta_3 &= \frac{\sin\theta_3^*}{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}.\\+\tan\theta_3 &= \frac{\sin\theta_3^*}{\delta_{23}^*+\cos\theta_3^*}.\\
 \end{align} \end{align}
  
行 326: 行 274:
 \begin{align} \begin{align}
 \cos\theta_3 \cos\theta_3
-&= \frac{1}{\sqrt{1+\left[\frac{\sin\theta_3^*}{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}\right]^2}}\\ +&= \frac{1}{\sqrt{1+\left[\frac{\sin\theta_3^*}{\delta_{23}^*+\cos\theta_3^*}\right]^2}}\\ 
-&= \frac{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}{\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}+&= \frac{\delta_{23}^*+\cos\theta_3^*}{\sqrt{\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}
 \end{align} \end{align}
  
行 335: 行 283:
  
 \begin{align} \begin{align}
-|\boldsymbol{\beta}_3^*| = |\boldsymbol{\beta}_1^*|\\+|\boldsymbol{v}_3^*| = |\boldsymbol{v}_1^*|\\
 \delta_{23}^* = \delta_{21}^* \delta_{23}^* = \delta_{21}^*
 \end{align} \end{align}
行 343: 行 291:
 \begin{align} \begin{align}
 \tan\theta_3 \tan\theta_3
-&= \frac{\sin\theta_3^*}{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}.\\ +&= \frac{\sin\theta_3^*}{\delta_{23}^*+\cos\theta_3^*}.\\ 
-&= \frac{\sin\theta_3^*}{\gamma_2^*\left(\delta_{21}^*+\cos\theta_3^*\right)}.\\+&= \frac{\sin\theta_3^*}{\delta_{21}^*+\cos\theta_3^*}.\\
 \end{align} \end{align}
  
行 350: 行 298:
  
 \begin{align} \begin{align}
-\begin{pmatrix} +|\boldsymbol{v}_4|\cos\theta_4 &= -|\boldsymbol{v}_4^*|\cos\theta_4^*+v_{\rm cm}\\ 
-E_4\\ +|\boldsymbol{v}_4|\sin\theta_4 &= |\boldsymbol{v}_4^*|\sin\theta_4^*
-|\boldsymbol{p}_4|\cos\theta_4 +
-\end{pmatrix} +
-&= +
-\begin{pmatrix} +
-\gamma_{\rm cm} & \beta_{\rm cm}\gamma_{\rm cm}\\ +
-\beta_{\rm cm}\gamma_{\rm cm} & \gamma_{\rm cm} +
-\end{pmatrix} +
-\begin{pmatrix} +
-E_4^*\\ +
--|\boldsymbol{p}_4^*|\cos\theta_4^* +
-\end{pmatrix}\\ +
-|\boldsymbol{p}_4|\cos\theta_4 &= \gamma_{\rm cm}\left(\beta_{\rm cm}E_4^*-|\boldsymbol{p}_4^*|\cos\theta_4^*\right)\\ +
-|\boldsymbol{p}_4|\sin\theta_4 &= |\boldsymbol{p}_4^*|\sin\theta_4^*+
 \end{align} \end{align}
  
行 370: 行 305:
  
 \begin{align} \begin{align}
-\frac{|\boldsymbol{p}_4|\sin\theta_4}{|\boldsymbol{p}_4|\cos\theta_4} &= \frac{|\boldsymbol{p}_4^*|\sin\theta_4^*}{\gamma_{\rm cm}\left(\beta_{\rm cm}E_4^*-|\boldsymbol{p}_4^*|\cos\theta_4\right)}\\ +\frac{|\boldsymbol{v}_4|\sin\theta_4}{|\boldsymbol{v}_4|\cos\theta_4} &= \frac{|\boldsymbol{v}_4^*|\sin\theta_4^*}{v_{\rm cm}-|\boldsymbol{v}_4^*|\cos\theta_4^*}\\ 
-\tan\theta_4 &= \frac{\sin\theta_4^*}{\gamma_{\rm cm}\left(\beta_{\rm cm}(E_4^*/|\boldsymbol{p}_4^*|)-\cos\theta_4^*\right)}\\ +\tan\theta_4 &= \frac{\sin\theta_4^*}{v_{\rm cm}/|\boldsymbol{v}_4^*|-\cos\theta_4^*}\\
-\tan\theta_4 &= \frac{\sin\theta_4^*}{\gamma_{\rm cm}\left(\beta_{\rm cm}/|\boldsymbol{\beta}_4^*|-\cos\theta_4^*\right)}\\+
 \end{align} \end{align}
  
-By using $\theta_4^* = \theta_3^*$, $\beta_{\rm cm} = |\boldsymbol{\beta}_2^*|$ and $\gamma_{\rm cm} = \gamma_2^*$,+By using $\theta_4^* = \theta_3^*$ and $v_{\rm cm} = |\boldsymbol{v}_2^*|$,
 \begin{align} \begin{align}
-\tan\theta_4 = \frac{\sin\theta_3^*}{\gamma_2^*\left(|\boldsymbol{\beta}_2^*|/|\boldsymbol{\beta}_4^*|-\cos\theta_3^*\right)}+\tan\theta_4 = \frac{\sin\theta_3^*}{|\boldsymbol{v}_2^*|/|\boldsymbol{v}_4^*|-\cos\theta_3^*}
 \end{align} \end{align}
  
-For elastic scattering, $m_2=m_4$, $|\boldsymbol{p}_2^*|=|\boldsymbol{p}_4^*|$, and $|\boldsymbol{\beta}_2^*|=|\boldsymbol{\beta}_4^*|$. Therefore,+For elastic scattering, $m_2=m_4$ and $|\boldsymbol{v}_2^*|=|\boldsymbol{v}_4^*|$. Therefore,
 \begin{align} \begin{align}
-\tan\theta_4 = \frac{\sin\theta_3^*}{\gamma_2^*(1-\cos\theta_3^*)}+\tan\theta_4 = \frac{\sin\theta_3^*}{1-\cos\theta_3^*}
 \end{align} \end{align}
  
行 388: 行 322:
  
 \begin{align} \begin{align}
-\tan\theta_4 = \frac{1}{\gamma_2^*}\cot\frac{\theta_3^*}{2}+\tan\theta_4 = \cot\frac{\theta_3^*}{2}
 \end{align} \end{align}
  
行 398: 行 332:
 \begin{align} \begin{align}
 \tan\theta_3 \tan\theta_3
-&= \frac{\sin\theta_3^*}{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}.\\ +&= \frac{\sin\theta_3^*}{\delta_{23}^*+\cos\theta_3^*}.\\ 
-&= \frac{\sin\theta_3^*}{\gamma_2^*\left(1+\cos\theta_3^*\right)}.\\+&= \frac{\sin\theta_3^*}{1+\cos\theta_3^*}.\\
 \end{align} \end{align}
  
行 407: 行 341:
 From an equation in the derivation of Quantity 7., From an equation in the derivation of Quantity 7.,
 \begin{align} \begin{align}
-\tan\theta_3 &= \frac{\sin\theta_3^*}{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}.\\+\tan\theta_3 &= \frac{\sin\theta_3^*}{\delta_{23}^*+\cos\theta_3^*}.\\
 \end{align} \end{align}
  
行 413: 行 347:
 \begin{align} \begin{align}
 \tan^2\theta_3 \tan^2\theta_3
-&= \frac{\sin^2\theta_3^*}{\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}\\ +&= \frac{\sin^2\theta_3^*}{\left(\delta_{23}^*+\cos\theta_3^*\right)^2}\\ 
-&= \frac{1-\cos^2\theta_3^*}{\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2},\\ +&= \frac{1-\cos^2\theta_3^*}{\left(\delta_{23}^*+\cos\theta_3^*\right)^2},\\ 
-&\Rightarrow (\gamma_2^*\tan\theta_3)^2\left(\delta_{23}^*+\cos\theta_3^*\right)^2=1-\cos^2\theta_3^*\\ +&\Rightarrow \tan^2\theta_3\left(\delta_{23}^*+\cos\theta_3^*\right)^2=1-\cos^2\theta_3^*\\ 
-&\Rightarrow \left[(\gamma_2^*\tan\theta_3)^2+1\right]\cos\theta_3^{*2}+2\delta_{23}^*(\gamma_2^*\tan\theta_3)^2\cos\theta_3^*+\delta_{23}^{*2}(\gamma_2^*\tan\theta_3)^2-1=0\\+&\Rightarrow \left[\tan^2\theta_3+1\right]\cos\theta_3^{*2}+2\delta_{23}^*(\tan^2\theta_3\cos\theta_3^*+\delta_{23}^{*2}\tan^2\theta_3-1=0\\
 \end{align} \end{align}
  
行 425: 行 359:
 is is
 \begin{align} \begin{align}
-x=\frac{b}{a}\pm\sqrt{\left(\frac{b}{a}\right)^2-\frac{c}{a}}.+x=-\frac{b}{a}\pm\sqrt{\left(\frac{b}{a}\right)^2-\frac{c}{a}}.
 \end{align} \end{align}
 Therefore, Therefore,
 \begin{align} \begin{align}
-\cos\theta_3^*=\frac{\delta_{23}^*(\gamma_2^*\tan\theta_3)^2}{(\gamma_2^*\tan\theta_3)^2+1}\pm\sqrt{\left(\frac{\delta_{23}^*(\gamma_2^*\tan\theta_3)^2}{(\gamma_2^*\tan\theta_3)^2+1}\right)^2-\frac{\delta_{23}^{*2}(\gamma_2^*\tan\theta_3)^2-1}{(\gamma_2^*\tan\theta_3)^2+1}}.+\cos\theta_3^*=-\frac{\delta_{23}^*\tan^2\theta_3}{\tan^2\theta_3+1}\pm\sqrt{\left(\frac{\delta_{23}^*\tan^2\theta_3}{\tan^2\theta_3+1}\right)^2-\frac{\delta_{23}^{*2}\tan^2\theta_3-1}{\tan^2\theta_3+1}}.
 \end{align} \end{align}
  
行 436: 行 370:
  
 \begin{align} \begin{align}
-\begin{pmatrix} +|\boldsymbol{v}_3^*|\cos\theta_{\rm cm} &= |\boldsymbol{v}_3|\cos\theta_{\rm lab}-v_{\rm cm}\\ 
-E_3^*\\ +|\boldsymbol{v}_3^*|\sin\theta_{\rm cm} &= |\boldsymbol{v}_3|\sin\theta_{\rm lab}
-|\boldsymbol{p}_3^*|\cos\theta_{\rm cm} +
-\end{pmatrix} +
-&= +
-\begin{pmatrix} +
-\gamma_{\rm cm} & -\beta_{\rm cm}\gamma_{\rm cm}\\ +
--\beta_{\rm cm}\gamma_{\rm cm} & \gamma_{\rm cm} +
-\end{pmatrix} +
-\begin{pmatrix} +
-E_3\\ +
-|\boldsymbol{p}_3|\cos\theta_{\rm lab} +
-\end{pmatrix}\\ +
-|\boldsymbol{p}_3^*|\cos\theta_{\rm cm} &= \gamma_{\rm cm}\left(|\boldsymbol{p}_3|\cos\theta_{\rm lab}-\beta_{\rm cm}E_3\right)\\ +
-|\boldsymbol{p}_3^*|\sin\theta_{\rm cm} &= |\boldsymbol{p}_3|\sin\theta_{\rm lab}+
 \end{align} \end{align}
  
行 456: 行 377:
  
 \begin{align} \begin{align}
-\frac{|\boldsymbol{p}_3^*|\sin\theta_{\rm cm}}{|\boldsymbol{p}_3^*|\cos\theta_{\rm cm}} &= \frac{|\boldsymbol{p}_3|\sin\theta_{\rm lab}}{\gamma_{\rm cm}\left(|\boldsymbol{p}_3|\cos\theta_{\rm lab}-\beta_{\rm cm}E_3\right)}\\ +\frac{|\boldsymbol{v}_3^*|\sin\theta_{\rm cm}}{|\boldsymbol{v}_3^*|\cos\theta_{\rm cm}} &= \frac{|\boldsymbol{v}_3|\sin\theta_{\rm lab}}{|\boldsymbol{v}_3|\cos\theta_{\rm lab}-v_{\rm cm}}\\ 
-\tan\theta_{\rm cm} &= \frac{\sin\theta_{\rm lab}}{\gamma_{\rm cm}\left(\cos\theta_{\rm lab}-\beta_{\rm cm}(E_3/|\boldsymbol{p}_3|)\right)}\\ +\tan\theta_{\rm cm} &= \frac{\sin\theta_{\rm lab}}{\cos\theta_{\rm lab}-v_{\rm cm}/|\boldsymbol{v}_3|}\\
-\tan\theta_{\rm cm} &= \frac{\sin\theta_{\rm lab}}{\gamma_{\rm cm}\left(\cos\theta_{\rm lab}-\beta_{\rm cm}/|\boldsymbol{\beta}_3|\right)}\\+
 \end{align} \end{align}
  
行 473: 行 393:
 By using the Quantity 7. and $(f/g)'=(f'g-fg')/g^2$, By using the Quantity 7. and $(f/g)'=(f'g-fg')/g^2$,
 \begin{align} \begin{align}
-\frac{d(\cos\theta_3)}{d(\cos\theta_3^*)} +\frac{d\Omega_3}{d\Omega_3^*}=\frac{d(\cos\theta_3)}{d(\cos\theta_3^*)} 
-&= \frac{d}{d(\cos\theta_3^*)}\left[\frac{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}{\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\right]\\ +&= \frac{d}{d(\cos\theta_3^*)}\left[\frac{\delta_{23}^*+\cos\theta_3^*}{\sqrt{\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\right]\\ 
-&= \left.\left[\frac{d\left[\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)\right]}{d(\cos\theta_3^*)}\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}-\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)\frac{d\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}{d(\cos\theta_3^*)}\right]\right/\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]\\ +&= \left.\left[\frac{d\left(\delta_{23}^*+\cos\theta_3^*\right)}{d(\cos\theta_3^*)}\sqrt{\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2}-\left(\delta_{23}^*+\cos\theta_3^*\right)\frac{d\sqrt{\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}{d(\cos\theta_3^*)}\right]\right/\left[\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]\\ 
-&= \left.\left[\gamma_2^*\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}-\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)\frac{1}{2\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\frac{d\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]}{d(\cos\theta_3^*)}\right]\right/\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]\\ +&= \left.\left[\frac{d\left(\delta_{23}^*+\cos\theta_3^*\right)}{d(\cos\theta_3^*)}\sqrt{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*}-\left(\delta_{23}^*+\cos\theta_3^*\right)\frac{d\sqrt{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*}}{d(\cos\theta_3^*)}\right]\right/\left[\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]\\ 
-&= \left.\left[\gamma_2^*\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}-\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)\frac{1}{2\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\frac{d\left[1-\cos^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^{*2}+2\delta_{23}^{*}\cos\theta_3^*+\cos^2\theta_3^*\right)\right]}{d(\cos\theta_3^*)}\right]\right/\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]\\ +&= \left.\left[\sqrt{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*}-\left(\delta_{23}^*+\cos\theta_3^*\right)\frac{1}{2\sqrt{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*}}\frac{d\left(1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*\right)}{d(\cos\theta_3^*)}\right]\right/\left[\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]\\ 
-&= \left.\left[\gamma_2^*\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}-\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)\frac{1}{2\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\frac{d\left[(\gamma_2^{*2}-1)\cos^2\theta_3^*+2\delta_{23}^{*}\gamma_2^{*2}\cos\theta_3^*+\delta_{23}^{*2}\gamma_2^{*2}+1\right]}{d(\cos\theta_3^*)}\right]\right/\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]\\ +&= \left.\left[\sqrt{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*}-\left(\delta_{23}^*+\cos\theta_3^*\right)\frac{1}{2\sqrt{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*}}\cdot 2\delta_{23}^{*}\right]\right/\left[\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]\\ 
-&= \left.\left[\gamma_2^*\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}-\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)\frac{1}{2\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\left[2(\gamma_2^{*2}-1)\cos\theta_3^*+2\delta_{23}^{*}\gamma_2^{*2}\right]\right]\right/\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]\\ +&= \left.\left[\left[1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*\right]-\delta_{23}^{*}\left(\delta_{23}^*+\cos\theta_3^*\right)\right]\right/\left[\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]^{3/2}\\ 
-&= \left.\left[\gamma_2^*\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]-\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)\left[(\gamma_2^{*2}-1)\cos\theta_3^*+\delta_{23}^{*}\gamma_2^{*2}\right]\right]\right/\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]^{3/2}\\ +&= \left.\left[1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*-\delta_{23}^{*2}-\delta_{23}^{*}\cos\theta_3^*\right]\right/\left[\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]^{3/2}\\ 
-&= \gamma_2^*\left.\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2-\left(\delta_{23}^*+\cos\theta_3^*\right)\left[(\gamma_2^{*2}-1)\cos\theta_3^*+\delta_{23}^{*}\gamma_2^{*2}\right]\right]\right/\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]^{3/2}\\ +&= \frac{1+\delta_{23}^{*}\cos\theta_3^*}{\left[\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]^{3/2}}
-&= \gamma_2^*\left.\left[(\gamma_2^{*2}-1)\cos^2\theta_3^*+2\delta_{23}^{*}\gamma_2^{*2}\cos\theta_3^*+\delta_{23}^{*2}\gamma_2^{*2}+1-\left[(\gamma_2^{*2}-1)\cos^2\theta_3^*+\delta_{23}^{*}\gamma_2^{*2}\cos\theta_3^*+\delta_{23}^{*}(\gamma_2^{*2}-1)\cos\theta_3^*+\delta_{23}^{*2}\gamma_2^{*2}\right]\right]\right/\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]^{3/2}\\ +
-&= \gamma_2^*\left.\left[(\gamma_2^{*2}-1)\cos^2\theta_3^*+2\delta_{23}^{*}\gamma_2^{*2}\cos\theta_3^*+\delta_{23}^{*2}\gamma_2^{*2}+1-\left[(\gamma_2^{*2}-1)\cos^2\theta_3^*+2\delta_{23}^{*}\gamma_2^{*2}\cos\theta_3^*-\delta_{23}^{*}\cos\theta_3^*+\delta_{23}^{*2}\gamma_2^{*2}\right]\right]\right/\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]^{3/2}\\ +
-&= \frac{\gamma_2^*(1+\delta_{23}^{*}\cos\theta_3^*)}{\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]^{3/2}}+
 \end{align} \end{align}
  
行 491: 行 408:
 From Quantity 7., From Quantity 7.,
 \begin{align} \begin{align}
-\cos\theta_3 &= \frac{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}{\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}.+\cos\theta_3 
 +&= \frac{\delta_{23}^*+\cos\theta_3^*}{\sqrt{\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}
 \end{align} \end{align}
 In general, $\sin\theta=\sqrt{1-\cos^2\theta}$. Therefore, In general, $\sin\theta=\sqrt{1-\cos^2\theta}$. Therefore,
 \begin{align} \begin{align}
 \sin\theta_3 \sin\theta_3
-&= \sqrt{1-\left[\frac{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}{\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\right]^2}\\ +&= \sqrt{1-\left[\frac{\delta_{23}^*+\cos\theta_3^*}{\sqrt{\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\right]^2}\\ 
-&= \sqrt{\frac{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2-\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\\ +&= \sqrt{\frac{\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2-\left(\delta_{23}^*+\cos\theta_3^*\right)^2}{\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\\ 
-&= \sqrt{\frac{\sin^2\theta_3^*}{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\\ +&= \sqrt{\frac{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*-\left(\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*+\cos^2\theta_3^*\right)}{\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\\ 
-&= \frac{\sin\theta_3^*}{\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\\+&= \sqrt{\frac{1-\cos^2\theta_3^*}{\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\\ 
 +&= \sqrt{\frac{\sin^2\theta_3^*}{\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\\ 
 +&= \frac{\sin\theta_3^*}{\sqrt{\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\\
 \Rightarrow \frac{\sin\theta_3}{\sin\theta_3^*} \Rightarrow \frac{\sin\theta_3}{\sin\theta_3^*}
-&= \frac{1}{\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}+&= \frac{1}{\sqrt{\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}
 \end{align} \end{align}
 By substituting this formula into the first formula of Quantity 9., By substituting this formula into the first formula of Quantity 9.,
 \begin{align} \begin{align}
 \frac{d\Omega_3}{d\Omega_3^*} \frac{d\Omega_3}{d\Omega_3^*}
-&= \frac{\gamma_2^*(1+\delta_{23}^*\cos\theta_3^*)}{\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]^{3/2}}\\ +&= \frac{1+\delta_{23}^*\cos\theta_3^*}{\left[\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]^{3/2}}\\ 
-&= \frac{\sin^3\theta_3}{\sin^3\theta_3^*}\gamma_2^*(1+\delta_{23}^*\cos\theta_3^*)\\+&= \frac{\sin^3\theta_3}{\sin^3\theta_3^*}(1+\delta_{23}^*\cos\theta_3^*)\\
 \end{align} \end{align}
  
行 516: 行 436:
  
 \begin{align} \begin{align}
-\frac{d(\cos\theta_3)}{d(\cos\theta_3^*)} +\frac{d\Omega_3}{d\Omega_3^*} 
-&= \frac{\gamma_2^*(1+\delta_{23}^{*}\cos\theta_3^*)}{\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]^{3/2}} +&= \frac{1+\delta_{23}^{*}\cos\theta_3^*}{\left[\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]^{3/2}}\\ 
-&= \frac{\gamma_2^*(1+\cos\theta_3^*)}{\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(1+\cos\theta_3^*\right)^2\right]^{3/2}}+&= \frac{1+\delta_{23}^{*}\cos\theta_3^*}{\left[1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*\right]^{3/2}}\\ 
 +&= \frac{1+\cos\theta_3^*}{\left[2+2\cos\theta_3^*\right]^{3/2}}\\ 
 +&= \frac{1+\cos\theta_3^*}{\left[2\left(1+\cos\theta_3^*\right)\right]^{3/2}}\\ 
 +&= \frac{1}{2^{3/2}\sqrt{1+\cos\theta_3^*}}
 \end{align} \end{align}
- 
-===  Derivation of Quantity 10. Relations between the $\gamma$ factors === 
- 
-** The first formula of the General Formulae ** 
- 
-In general, 
- 
-\begin{align} 
-|\boldsymbol{p}|=m\gamma|\boldsymbol{\beta}|=m\sqrt{\gamma^2-1}. 
-\end{align} 
-On the other hand, in the Center of mass system, 
-\begin{align} 
-|\boldsymbol{p}_1^*|=|\boldsymbol{p}_2^*|. 
-\end{align} 
-Therefore, 
-\begin{align} 
-m_1\sqrt{\gamma_1^{*2}-1}=m_2\sqrt{\gamma_2^{*2}-1}\\ 
-\Rightarrow m_1^2(\gamma_1^{*2}-1)=m_2^2(\gamma_2^{*2}-1)\\ 
-\Rightarrow (\gamma_1^{*2}-1)=k_{21}^2(\gamma_2^{*2}-1). 
-\end{align} 
- 
-** The third formula of the General Formulae ** 
- 
-From Quantity 1. and 5., 
- 
-\begin{align} 
-\gamma_2^* 
-&= \gamma_\mathrm{cm}\\ 
-&= \frac{E_1+m_2}{W}\\ 
-&= \frac{E_1+m_2}{\sqrt{m_1^2+m_2^2+2m_2E_1}}\\ 
-&= \frac{m_1\gamma_1+m_2}{\sqrt{m_1^2+m_2^2+2m_1m_2\gamma_1}}. 
-\end{align} 
-Then, by using $k_{12}=m_1/m_2$, 
- 
-\begin{align} 
-\gamma_2^* = \frac{k_{21}+\gamma_1}{\sqrt{1+k_{21}^2+2\gamma_1k_{21}}} 
-\end{align} 
- 
- 
-** The second formula of the General Formulae ** 
- 
-By substituting the third formula into the first formula, 
- 
-\begin{align} 
-(\gamma_1^{*2}-1) 
-&= k_{21}^2\left[\frac{(k_{21}+\gamma_1)^2}{1+k_{21}^2+2\gamma_1k_{21}}-1\right]\\ 
-&= k_{21}^2\left[\frac{k_{21}^2+\gamma_1^2+2\gamma_1k_{21}}{1+k_{21}^2+2\gamma_1k_{21}}-1\right]\\ 
-&= k_{21}^2\frac{k_{21}^2+\gamma_1^2+2\gamma_1k_{21}-(1+k_{21}^2+2\gamma_1k_{21})}{1+k_{21}^2+2\gamma_1k_{21}}\\ 
-&= k_{21}^2\frac{\gamma_1^2-1}{1+k_{21}^2+2\gamma_1k_{21}}\\ 
-&= \frac{\gamma_1^2-1}{(1/k_{21})^2+1+2\gamma_1(1/k_{21})}\\ 
-&= \frac{\gamma_1^2-1}{1+k_{12}^2+2\gamma_1k_{12}}\\ 
-\Leftrightarrow 
-\gamma_1^{*2} 
-&= \frac{\gamma_1^2-1}{1+k_{12}^2+2\gamma_1k_{12}}+1\\ 
-&= \frac{\gamma_1^2-1+(1+k_{12}^2+2\gamma_1k_{12})}{1+k_{12}^2+2\gamma_1k_{12}}\\ 
-&= \frac{\gamma_1^2+k_{12}^2+2\gamma_1k_{12}}{1+k_{12}^2+2\gamma_1k_{12}}\\ 
-&= \frac{(k_{12}+\gamma_1)^2}{1+k_{12}^2+2\gamma_1k_{12}}\\ 
-\end{align} 
- 
-Then, 
- 
-\begin{align} 
-\gamma_1^* = \frac{k_{12}+\gamma_1}{\sqrt{1+k_{12}^2+2\gamma_1k_{12}}} 
-\end{align} 
- 
-** The formulae for N-N Scattering ** 
- 
-For N-N scattering, $m_1 = m_2 = m_\mathrm{N}$. Therefore, 
- 
-\begin{align} 
-k_{12} 
-&= \frac{m_1}{m_2}\\ 
-&= \frac{m_\mathrm{N}}{m_\mathrm{N}}\\ 
-&= 1\\ 
-\end{align} 
- 
-Similarly, $k_{21} = 1$. As a result, from the first formula of the General Formulae, 
- 
-\begin{align} 
-(\gamma_1^{*2}-1)=k_{21}^2(\gamma_2^{*2}-1)\\ 
-\Rightarrow \gamma_1^{*2}-1=\gamma_2^{*2}-1\\ 
-\Rightarrow \gamma_1^*=\gamma_2^*. 
-\end{align} 
- 
-And, from the second formula of the General Formulae, 
- 
-\begin{align} 
-\gamma_1^* 
-&= \frac{k_{12}+\gamma_1}{\sqrt{1+k_{12}^2+2\gamma_1k_{12}}}\\ 
-&= \frac{1+\gamma_1}{\sqrt{1+1+2\gamma_1}}\\ 
-&= \frac{1+\gamma_1}{\sqrt{2(1+\gamma_1)}}\\ 
-&= \sqrt{\frac{1+\gamma_1}{2}}\\ 
-\end{align} 
- 
  
 === Derivation of Quantity 11. Lab quantity relations === === Derivation of Quantity 11. Lab quantity relations ===
行 640: 行 469:
 \begin{align} \begin{align}
 2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3 2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3
-&= \boldsymbol{p}_1^2 + \boldsymbol{p}_3^2 - \boldsymbol{p}_4^2\\ +&= \boldsymbol{p}_1^2 + \boldsymbol{p}_3^2 - \boldsymbol{p}_4^2.
-&= E_1^2 - m_1^2 + E_3^2 - m_3^2 - E_4^2 + m_4^2.\\ +
-\end{align} +
-From the law of conservation of energy, +
-\begin{align} +
-E_1+m_2&=E_3+E_4,\\ +
-E_4&=E_1+m_2-E_3,\\ +
-E_4^2&=E_1^2+m_2^2+E_3^2-2(E_1+m_2)E_3+2E_1m_2.\\ +
-\end{align} +
-By substituting this formula into the above formula, +
-\begin{align} +
-2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3 +
-&= m_4^2 - m_1^2 - m_2^2 - m_3^2+2(E_1+m_2)E_3-2E_1m_2.\\+
 \end{align} \end{align}
  
行 666: 行 483:
 2|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4) 2|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4)
  &= 2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3-2\boldsymbol{p}_3^2\\  &= 2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3-2\boldsymbol{p}_3^2\\
- &= 2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3-2E_3^2+2m_3^2.\\ 
 \end{align} \end{align}
 By substituting the first formula of the Quantity 11. Lab quantity relations into this formula, By substituting the first formula of the Quantity 11. Lab quantity relations into this formula,
 \begin{align} \begin{align}
 2|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4) 2|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4)
-&= m_4^2 - m_1^2 - m_2^2 - m_3^2+2(E_1+m_2)E_3-2E_1m_2 -2E_3^2+2m_3^2\\ +&= \boldsymbol{p}_1^2 + \boldsymbol{p}_3^2 - \boldsymbol{p}_4^2 -2\boldsymbol{p}_3^2\\ 
-&= m_3^2 + m_4^2 - m_1^2 - m_2^2+2(E_1+m_2)E_3-2E_1m_2 -2E_3^2\\+&= \boldsymbol{p}_1^2 - \boldsymbol{p}_3^2 - \boldsymbol{p}_4^2
 \end{align} \end{align}
-By substituting the $E_1+m_2=E_3+E_4$ into the formula,+ 
 +** The formula for N-N Scattering ** 
 + 
 +From the law of conservation of energy, 
 \begin{align} \begin{align}
-2|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4) +\frac{\boldsymbol{p}_1^2}{2m_1} = \frac{\boldsymbol{p}_3^2}{2m_3} + \frac{\boldsymbol{p}_4^2}{2m_4}
-&= m_3^2 + m_4^2 - m_1^2 - m_2^2+2(E_3+E_4)E_3-2E_1m_2 -2E_3^2\\ +
-&= m_3^2 + m_4^2 - m_1^2 - m_2^2-2E_1m_2+2E_3E_4\\+
 \end{align} \end{align}
  
-** The formula for N-N Scattering **+For N-N scattering, $m_1 = m_2 = m_3 = m_4 = m_\mathrm{N}$. Therefore, 
  
-For N-N scattering, $m_1 = m_2 = m_3 = m_4 = m_\mathrm{N}$. Therefore, the second formula of the General Formulae becomes+\begin{align} 
 +\boldsymbol{p}_1^2 = \boldsymbol{p}_3^2 + \boldsymbol{p}_4^2. 
 +\end{align}
  
 +From this formula and the General formula,
 \begin{align} \begin{align}
 2|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4) 2|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4)
-&= m_3^2 + m_4^2 - m_1^2 - m_2^2+2(E_3+E_4)E_3-2E_1m_2 -2E_3^2\\ +&= \boldsymbol{p}_1^2 - \boldsymbol{p}_3^2 - \boldsymbol{p}_4^2\\ 
-&= m_\mathrm{N}^2 + m_\mathrm{N}^2 - m_\mathrm{N}^2 - m_\mathrm{N}^2-2E_1m_\mathrm{N}+2E_3E_4\\ +&= 0
-&= -2E_1m_\mathrm{N}+2E_3E_4\\ +
-\Rightarrow +
-|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4) +
-&= -E_1m_\mathrm{N}+E_3E_4\\ +
-&= E_3E_4-m_\mathrm{N}(E_1+m_\mathrm{N})+m_\mathrm{N}^2\\+
 \end{align} \end{align}
  
-From the law of conservation of energy, $E_1+m_\mathrm{N}=E_3+E_4$. Therefore,+Therefore, 
 \begin{align} \begin{align}
-|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4) +\cos(\theta_3+\theta_4) &= 0\\ 
-&= E_3E_4-m_\mathrm{N}(E_1+m_\mathrm{N})+m_\mathrm{N}^2\\ +\theta_3+\theta_4 &= 90^\circ.
-&= E_3E_4-m_\mathrm{N}(E_3+E_4)+m_\mathrm{N}^2\\ +
-&= (E_3-m_\mathrm{N})(E_4-m_\mathrm{N})\\ +
-&= T_3T_4+
 \end{align} \end{align}
- 
  
 === Derivation of Quantity 12. Maximum K.E. transfer to a stationary particle === === Derivation of Quantity 12. Maximum K.E. transfer to a stationary particle ===
  
-** The first formula of the General Formulae **+** The General Formula ** 
 \begin{align} \begin{align}
-\begin{pmatrix} +|\boldsymbol{v}_4|\cos\theta_4 &= -|\boldsymbol{v}_4^*|\cos\theta_4^*+v_{\rm cm}\\ 
-E_4\\ +|\boldsymbol{v}_4|\sin\theta_4 &= |\boldsymbol{v}_4^*|\sin\theta_4^*
-|\boldsymbol{p}_4|\cos\theta_\mathrm{lab} +
-\end{pmatrix} +
-&= +
-\begin{pmatrix} +
-\gamma_\mathrm{cm} & \beta_\mathrm{cm}\gamma_\mathrm{cm}\\ +
-\beta_\mathrm{cm}\gamma_\mathrm{cm} & \gamma_\mathrm{cm} +
-\end{pmatrix} +
-\begin{pmatrix} +
-E_4^*\\ +
--|\boldsymbol{p}_4^*|\cos\theta_\mathrm{cm} +
-\end{pmatrix}\\ +
-\Rightarrow E_4 &=  \gamma_\mathrm{cm} (E_4^* - \beta_\mathrm{cm}|\boldsymbol{p}_4^*|\cos\theta_\mathrm{cm})+
 \end{align} \end{align}
  
-Therefore, $E_4$ becomes maximum at $\theta_\mathrm{cm}=\pi$. The maximum value is+By taking the squares for the both sides, 
 \begin{align} \begin{align}
-E_\mathrm{4max} &=  \gamma_\mathrm{cm} (E_4^* + \beta_\mathrm{cm}|\boldsymbol{p}_4^*|).+|\boldsymbol{v}_4|^2\cos^2\theta_4 &= |\boldsymbol{v}_4^*|^2\cos^2\theta_4^*+v_{\rm cm}^2-2v_{\rm cm}|\boldsymbol{v}_4^*|\cos\theta_4^*\\ 
 +|\boldsymbol{v}_4|^2\sin^2\theta_4 &= |\boldsymbol{v}_4^*|^2\sin^2\theta_4^*.
 \end{align} \end{align}
  
-From this formula,+By adding each side, 
 \begin{align} \begin{align}
-T_\mathrm{max} +|\boldsymbol{v}_4|^2(\sin^2\theta_4+\cos^2\theta_4) &= |\boldsymbol{v}_4^*|^2(\sin^2\theta_4^*+\cos^2\theta_4^*)+v_{\rm cm}^2-2v_{\rm cm}|\boldsymbol{v}_4^*|\cos\theta_4^*\\ 
-&= E_\mathrm{4max} - m_4\\ +\Rightarrow |\boldsymbol{v}_4|^2 &= |\boldsymbol{v}_4^*|^2+v_{\rm cm}^2-2v_{\rm cm}|\boldsymbol{v}_4^*|\cos\theta_4^*
-&= \gamma_\mathrm{cm} (E_4^* + \beta_\mathrm{cm}|\boldsymbol{p}_4^*|) - m_4.+
 \end{align} \end{align}
  
-For the elastic scattering, $m_4 = m_2$, $|\boldsymbol{p}_4^*| = |\boldsymbol{p}_2^*|$, and $|E_4^*| = |E_2^*|$. Then,+Therefore, $|\boldsymbol{v}_4|$ becomes maximum at $\theta_4^* = \theta_\mathrm{cm}=180^\circ$. In this case, $\theta_4 = 0^\circ$ and $\boldsymbol{v}_4$ has the same direction as $\boldsymbol{v}_1$. From the law of conservation of energy and momentum, 
 \begin{align} \begin{align}
-T_\mathrm{max} +T_1 &= T_3 + T_4\\ 
-&= \gamma_\mathrm{cm} (E_2^* + \beta_\mathrm{cm}|\boldsymbol{p}_2^*|) - m_2\\ +\boldsymbol{p}_1 &= \boldsymbol{p}_3 + \boldsymbol{p}_4.
-&= \gamma_\mathrm{cm} [m_2\gamma_2^* + \beta_\mathrm{cm}(m_2|\boldsymbol{\beta}_2^*|\gamma_2^*)] - m_2.+
 \end{align} \end{align}
-By using $|\boldsymbol{\beta}_2^*| = \beta_\mathrm{cm}$ and $\gamma_2^* = \gamma_\mathrm{cm}$,+ 
 +In general, $T = \frac{\boldsymbol{p}^2}{2m}$. Therefore, the first equation is written as
 \begin{align} \begin{align}
-T_\mathrm{max} +\frac{\boldsymbol{p}_1^2}{2m_1} &= \frac{\boldsymbol{p}_3^2}{2m_3} + \frac{\boldsymbol{p}_4^2}{2m_4}\\ 
-&= \gamma_\mathrm{cm} [m_2\gamma_\mathrm{cm} + \beta_\mathrm{cm}(m_2\beta_\mathrm{cm}\gamma_\mathrm{cm})] - m_2\\ +\Rightarrow \frac{\boldsymbol{p}_1^2}{m_1} &= \frac{\boldsymbol{p}_3^2}{m_3} + \frac{\boldsymbol{p}_4^2}{m_4}
-&= m_2\gamma_\mathrm{cm}^2 (1+\beta_\mathrm{cm}^2) - m_2.\\+
 \end{align} \end{align}
-In general, $|\boldsymbol{\beta}| = \sqrt{\gamma^2-1}/\gamma$. Therefore,+ 
 +By substituting $\boldsymbol{p}_3 = \boldsymbol{p}_1 - \boldsymbol{p}_4$ into this formula,  
 \begin{align} \begin{align}
-T_\mathrm{max} +\frac{\boldsymbol{p}_1^2}{m_1} 
-&= m_2\gamma_\mathrm{cm}^2 \left(1+\frac{\gamma_\mathrm{cm}^2-1}{\gamma_\mathrm{cm}^2}\right) - m_2\\ +&= \frac{(\boldsymbol{p}_1 - \boldsymbol{p}_4)^2}{m_3} + \frac{\boldsymbol{p}_4^2}{m_4}\\ 
-&= m_2 (\gamma_\mathrm{cm}^2+\gamma_\mathrm{cm}^2-1) - m_2\\ +&= \frac{\boldsymbol{p}_1^2 - 2|\boldsymbol{p}_1||\boldsymbol{p}_4|\cos\theta_4 + \boldsymbol{p}_4^2}{m_3} + \frac{\boldsymbol{p}_4^2}{m_4}\\ 
-&= 2m_2 (\gamma_\mathrm{cm}^2-1)\\ +&= \frac{\boldsymbol{p}_1^2 - 2|\boldsymbol{p}_1||\boldsymbol{p}_4| + \boldsymbol{p}_4^2}{m_3} + \frac{\boldsymbol{p}_4^2}{m_4}.
-&= 2m_2 \boldsymbol{\beta}_\mathrm{cm}^2\gamma_\mathrm{cm}^2.\\+
 \end{align} \end{align}
-By using Quantity 4. and 5.,+ 
 +Therefore, by using $\boldsymbol{p}^2=|\boldsymbol{p}|^2$, 
 \begin{align} \begin{align}
-T_\mathrm{max} +\left(\frac{1}{m_3}+\frac{1}{m_4}\right)|\boldsymbol{p}_4|^2-\frac{2|\boldsymbol{p}_1|}{m_3}|\boldsymbol{p}_4|+\left(\frac{1}{m_3}-\frac{1}{m_1}\right)|\boldsymbol{p}_1|^2=0\\
-&= 2m_2 \left(\frac{\boldsymbol{p}_1}{E_1+m_2}\right)^2\left(\frac{E_1+m_2}{W}\right)^2\\ +
-&= 2m_2  \frac{\boldsymbol{p}_1^2}{W^2}.\\+
 \end{align} \end{align}
  
-$W$ can be written as+In general, the solution of the equation $ax^2-2bx+c=0$ is $x=\frac{b\pm\sqrt{b^2-ac}}{a}$. Therefore,
 \begin{align} \begin{align}
-W +|\boldsymbol{p}_4| 
-&= \sqrt{m_1^2+m_2^2+2m_2E_1}\\ +&=\frac{\frac{|\boldsymbol{p}_1|}{m_3}\pm\sqrt{\frac{|\boldsymbol{p}_1|^2}{m_3^2}-\left(\frac{1}{m_3}+\frac{1}{m_4}\right)\left(\frac{1}{m_3}-\frac{1}{m_1}\right)|\boldsymbol{p}_1|^2}}{\frac{1}{m_3}+\frac{1}{m_4}}\\ 
-&= \sqrt{m_1^2+m_2^2+2m_2m_1\gamma_1}\\ +&=\frac{\frac{1}{m_3}\pm\sqrt{\frac{1}{m_3^2}-\left(\frac{1}{m_3}+\frac{1}{m_4}\right)\left(\frac{1}{m_3}-\frac{1}{m_1}\right)}}{\frac{1}{m_3}+\frac{1}{m_4}}|\boldsymbol{p}_1|\\ 
-&= \sqrt{m_1m_2\left(\frac{m_1}{m_2}+\frac{m_2}{m_1}+2\gamma_1\right)}\\ +&=\frac{\frac{1}{m_3}\pm\sqrt{\frac{1}{m_3^2}-\left(\frac{1}{m_3^2}-\frac{1}{m_1m_3}-\frac{1}{m_1m_4}+\frac{1}{m_3m_4}\right)}}{\frac{1}{m_3}+\frac{1}{m_4}}|\boldsymbol{p}_1|\\ 
-&= \sqrt{m_1m_2\left(k_{12}+k_{21}+2\gamma_1\right)}.\\+&=\frac{\frac{1}{m_3}\pm\sqrt{\frac{1}{m_1m_3}+\frac{1}{m_1m_4}-\frac{1}{m_3m_4}}}{\frac{1}{m_3}+\frac{1}{m_4}}|\boldsymbol{p}_1|\\ 
 +&=\frac{\frac{1}{m_3}\pm\sqrt{\frac{1}{m_1m_3}+\frac{1}{m_1m_4}-\frac{1}{m_3m_4}}}{\frac{m_3+m_4}{m_3m_4}}|\boldsymbol{p}_1|\\ 
 +&=\frac{m_4\pm\sqrt{m_3m_4}\sqrt{\frac{m_3}{m_1}+\frac{m_4}{m_1}-1}}{m_3+m_4}|\boldsymbol{p}_1|\\ 
 +&=\frac{m_4\pm\sqrt{\frac{m_3m_4}{m_1}}\sqrt{m_3+m_4-m_1}}{m_3+m_4}|\boldsymbol{p}_1|\\
 \end{align} \end{align}
-By substituting this formula into the former formula,+ 
 +The direction of $\boldsymbol{p}_4$ is the same as $\boldsymbol{p}_1$, then
 \begin{align} \begin{align}
-T_\mathrm{max} +\boldsymbol{p}_4 
-&= 2m_2  \frac{\boldsymbol{p}_1^2}{m_1m_2\left(k_{12}+k_{21}+2\gamma_1\right)}\\ +&=\frac{m_4\pm\sqrt{\frac{m_3m_4}{m_1}}\sqrt{m_3+m_4-m_1}}{m_3+m_4}\boldsymbol{p}_1.
-&= \frac{2\boldsymbol{p}_1^2}{m_1(k_{12}+k_{21}+2\gamma_1)}.\\+
 \end{align} \end{align}
  
-** The second formula of the General Formulae ** +From $\boldsymbol{p}_3=\boldsymbol{p}_1-\boldsymbol{p}_4$,
- +
-From the first formula,+
 \begin{align} \begin{align}
-T_\mathrm{max} +\boldsymbol{p}_3 
-&= \frac{2\boldsymbol{p}_1^2}{m_1(k_{12}+k_{21}+2\gamma_1)}\\ +&= \boldsymbol{p}_1-\boldsymbol{p}_4\\ 
-&= \frac{2m_1^2\boldsymbol{\beta}_1^2\gamma_1^2}{m_1(k_{12}+k_{21}+2\gamma_1)}\\ +&= \boldsymbol{p}_1-\frac{m_4\pm\sqrt{\frac{m_3m_4}{m_1}(m_3+m_4-m_1)}}{m_3+m_4}\boldsymbol{p}_1\\ 
-&= \frac{2m_1\boldsymbol{\beta}_1^2\gamma_1^2}{k_{12}+k_{21}+2\gamma_1}.\\+&= \frac{m_3+m_4-m_4\mp\sqrt{\frac{m_3m_4}{m_1}(m_3+m_4-m_1)}}{m_3+m_4}\boldsymbol{p}_1\\ 
 +&= \frac{m_3\mp\sqrt{\frac{m_3m_4}{m_1}(m_3+m_4-m_1)}}{m_3+m_4}\boldsymbol{p}_1\\
 \end{align} \end{align}
-By using $m_1=m_2/k_{21}$,+ 
 +By using $T_4=\frac{|\boldsymbol{p}_4|^2}{2m_4}$ and $T_1=\frac{|\boldsymbol{p}_1|^2}{2m_1}$, 
 \begin{align} \begin{align}
-T_\mathrm{max} +T_4 
-&= \frac{2m_2\boldsymbol{\beta}_1^2\gamma_1^2}{k_{21}(k_{12}+k_{21}+2\gamma_1)}\\ +&=\frac{|\boldsymbol{p}_4|^2}{2m_4}\\ 
-&= \frac{2m_2\boldsymbol{\beta}_1^2\gamma_1^2}{1+k_{21}^2+2\gamma_1k_{21}}.\\+&=\frac{1}{2m_4}\frac{\left[m_4\pm\sqrt{\frac{m_3m_4}{m_1}}\sqrt{m_3+m_4-m_1}\right]^2}{(m_3+m_4)^2}|\boldsymbol{p}_1|^2\\ 
 +&=\frac{m_1}{m_4}\frac{\left[m_4\pm\sqrt{\frac{m_3m_4}{m_1}}\sqrt{m_3+m_4-m_1}\right]^2}{(m_3+m_4)^2}\frac{|\boldsymbol{p}_1|^2}{2m_1}\\ 
 +&=\frac{m_1}{m_4}\frac{\left[m_4\pm\sqrt{\frac{m_3m_4}{m_1}}\sqrt{m_3+m_4-m_1}\right]^2}{(m_3+m_4)^2}T_1\\ 
 +&=\frac{\left[\sqrt{\frac{m_1}{m_4}}m_4\pm\sqrt{\frac{m_1}{m_4}}\sqrt{\frac{m_3m_4}{m_1}}\sqrt{m_3+m_4-m_1}\right]^2}{(m_3+m_4)^2}T_1\\ 
 +&=\frac{\left[\sqrt{m_1m_4}\pm\sqrt{m_3(m_3+m_4-m_1)}\right]^2}{(m_3+m_4)^2}T_1.
 \end{align} \end{align}
-If $k_{21} = m2/m1 \ll 1$,+ 
 +As a result, 
 \begin{align} \begin{align}
-T_\mathrm{max} +T_\mathrm{max} = T_4 &=\frac{\left[\sqrt{m_1m_4}\pm\sqrt{m_3(m_3+m_4-m_1)}\right]^2}{(m_3+m_4)^2}T_1.
-&= \frac{2m_2\boldsymbol{\beta}_1^2\gamma_1^2}{1+k_{21}^2+2\gamma_1k_{21}}\\ +
-&\approx 2m_2\boldsymbol{\beta}_1^2\gamma_1^2.\\+
 \end{align} \end{align}
  
-** Another derivation of the second formula of the General Formulae **+By the way, from $T_3=T_1-T_4$, $T_3$ is  
  
-If $m2/m1 \ll 1$, $|\boldsymbol{\beta}_\mathrm{cm}| \approx |\boldsymbol{\beta}_1|$ and $\gamma_\mathrm{cm} \approx \gamma_1$. 
-Therefore, from the formula above, 
 \begin{align} \begin{align}
-T_\mathrm{max} +T_3 
-&= 2m_2 \boldsymbol{\beta}_\mathrm{cm}^2\gamma_\mathrm{cm}^2\\ +&= T_1-T_4\\ 
-&\approx 2m_2\boldsymbol{\beta}_1^2\gamma_1^2.\\+&= T_1-\frac{\left[\sqrt{m_1m_4}\pm\sqrt{m_3(m_3+m_4-m_1)}\right]^2}{(m_3+m_4)^2}T_1\\ 
 +&= T_1-\frac{\sqrt{m_1m_4}^2+\sqrt{m_3(m_3+m_4-m_1)}^2\pm2\sqrt{m_1m_4}\sqrt{m_3(m_3+m_4-m_1)}}{(m_3+m_4)^2}T_1\\ 
 +&= T_1-\frac{m_1m_4+m_3(m_3+m_4-m_1)\pm2\sqrt{m_1m_4}\sqrt{m_3(m_3+m_4-m_1)}}{(m_3+m_4)^2}T_1\\ 
 +&= T_1-\frac{m_1m_4-m_1m_3+m_3^2+m_3m_4\pm2\sqrt{m_1m_4}\sqrt{m_3(m_3+m_4-m_1)}}{(m_3+m_4)^2}T_1\\ 
 +&= \frac{(m_3+m_4)^2 -m_1m_4+m_1m_3-m_3^2-m_3m_4\mp2\sqrt{m_1m_4}\sqrt{m_3(m_3+m_4-m_1)}}{(m_3+m_4)^2}T_1\\ 
 +&= \frac{m_3^2+2m_3m_4+m_4^2-m_1m_4+m_1m_3-m_3^2-m_3m_4\mp2\sqrt{m_1m_4}\sqrt{m_3(m_3+m_4-m_1)}}{(m_3+m_4)^2}T_1\\ 
 +&= \frac{m_1m_3+m_3m_4+m_4^2-m_1m_4\mp2\sqrt{m_1m_4}\sqrt{m_3(m_3+m_4-m_1)}}{(m_3+m_4)^2}T_1\\ 
 +&= \frac{m_1m_3+m_4(m_3+m_4-m_1)\mp2\sqrt{m_1m_4}\sqrt{m_3(m_3+m_4-m_1)}}{(m_3+m_4)^2}T_1\\ 
 +&= \frac{\sqrt{m_1m_3}^2+\sqrt{m_4(m_3+m_4-m_1)}^2\mp2\sqrt{m_1m_4}\sqrt{m_3(m_3+m_4-m_1)}}{(m_3+m_4)^2}T_1\\ 
 +&= \frac{\left[\sqrt{m_1m_3}\mp\sqrt{m_4(m_3+m_4-m_1)}\right]^2}{(m_3+m_4)^2}T_1
 \end{align} \end{align}
  
 +** The first formula for elastic scattering **
  
-==== General memo ==== +For elastic scattering, $m_3 = m_1$ and $m_4 = m_2$. Therefore,
-=== Lorentz transformation ===+
 \begin{align} \begin{align}
-E^*&=\gamma_0(E-\boldsymbol{p}\cdot\boldsymbol{\beta}_0)\\ +\boldsymbol{p}_4&=\frac{2m_2}{m_1+m_2}\boldsymbol{p}_1\\ 
-\boldsymbol{p}^*&=\boldsymbol{p}+\boldsymbol{\beta}_0\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\boldsymbol{\beta}_0\cdot\boldsymbol{p}-E\right]+\boldsymbol{p}_3&=\frac{m_1-m_2}{m_1+m_2}\boldsymbol{p}_1\\ 
 +T_4&=\frac{4m_1m_2}{(m_1+m_2)^2}T_1\\ 
 +&=\frac{2m_2\boldsymbol{p}_1^2}{(m_1+m_2)^2}\\ 
 +&=2m_2\boldsymbol{v}_\mathrm{cm}\\ 
 +T_3&=\frac{(m_1- m_2)^2}{(m_1+m_2)^2}T_1
 \end{align} \end{align}
-== Beam axis == +As a result, 
-$\boldsymbol{p}$ can be written as+
 \begin{align} \begin{align}
-\boldsymbol{p} = p_{||}\boldsymbol{e}_{||} +p_{T}\boldsymbol{e}_T,+T_\mathrm{max}=T_4=\frac{4m_1m_2}{(m_1+m_2)^2}T_1
 \end{align} \end{align}
-where $\boldsymbol{e}_{||}$ ($\boldsymbol{e}_T$) is a unit vector parallel (perpendicular) to $\boldsymbol{\beta}_0$, and $p_{||}$ ($p_T$) is the component of $\boldsymbol{p}$ parallel (perpendicular) to $\boldsymbol{\beta}_0$. Similarly,+ 
 + 
 + 
 +** Another derivation of the first formula for elastic scattering ** 
 + 
 +From the above discussion, $|\boldsymbol{v}_4|$ becomes maximum at $\theta_4^* = \theta_\mathrm{cm}=180^\circ$. In this case, $\theta_4 = 0^\circ$ and $\boldsymbol{v}_4$ has the same direction as $\boldsymbol{v}_1$. For $\theta_4 = 0^\circ$, from the law of conservation of energy and momentum, 
 \begin{align} \begin{align}
-\boldsymbol{p}^* = p_{||}^*\boldsymbol{e}_{||} +p_{T}^*\boldsymbol{e}_T.+\frac{1}{2}m_1\boldsymbol{v}_1^2 = \frac{1}{2}m_1\boldsymbol{v}_3^2 + \frac{1}{2}m_2\boldsymbol{v}_4^2\\ 
 +m_1\boldsymbol{v}_1 = m_1\boldsymbol{v}_3 + m_2\boldsymbol{v}_4\\
 \end{align} \end{align}
-From the equation of $\boldsymbol{p}$,+ 
 +From the second equation,
 \begin{align} \begin{align}
-\boldsymbol{\beta}_0\cdot\boldsymbol{p} +\boldsymbol{v}_3 = \frac{m_1\boldsymbol{v}_1 - m_2\boldsymbol{v}_4}{m_1}.
- &= (\beta_0 \boldsymbol{e}_{||}) \cdot (p_{||}\boldsymbol{e}_{||} +p_{T}\boldsymbol{e}_T)\\ +
- &= \beta_0 p_{||} \boldsymbol{e}_{||} \cdot \boldsymbol{e}_{||} + \beta_0 p_{T} \boldsymbol{e}_{||} \cdot \boldsymbol{e}_T\\ +
- &= \beta_0 p_{||}.+
 \end{align} \end{align}
  
-By using these equations,+By substituting this formula into the formula of the law of conservation of energy, 
 \begin{align} \begin{align}
-E^* +\frac{1}{2}m_1\boldsymbol{v}_1^2 
- &=\gamma_0(E-\boldsymbol{p}\cdot\boldsymbol{\beta}_0)\\ +&= \frac{1}{2}m_1\frac{(m_1\boldsymbol{v}_1 - m_2\boldsymbol{v}_4)^2}{m_1^2} + \frac{1}{2}m_2\boldsymbol{v}_4^2\\ 
- &=\gamma_0(E-p_{||}\beta_0)\\ +&= \frac{1}{2}m_1\frac{m_1^2\boldsymbol{v}_1^2 + m_2^2\boldsymbol{v}_4^2 - 2m_1m_2|\boldsymbol{v}_1||\boldsymbol{v}_4|}{m_1^2} + \frac{1}{2}m_2\boldsymbol{v}_4^2\\ 
-\boldsymbol{p}^*&=\boldsymbol{p}+\boldsymbol{\beta}_0\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\boldsymbol{\beta}_0\cdot\boldsymbol{p}-E\right]\\ +\Rightarrow 0 &= \frac{1}{2}m_1\frac{m_2^2\boldsymbol{v}_4^2 - 2m_1m_2|\boldsymbol{v}_1||\boldsymbol{v}_4|}{m_1^2} + \frac{1}{2}m_2\boldsymbol{v}_4^2 \\ 
-p_{||}^*\boldsymbol{e}_{||} +p_{T}^*\boldsymbol{e}_T +\Rightarrow 0 &= \frac{m_2|\boldsymbol{v}_4| - 2m_1|\boldsymbol{v}_1|}{m_1} + |\boldsymbol{v}_4| \\ 
- &= p_{||}\boldsymbol{e}_{||} +p_{T}\boldsymbol{e}_T + \beta_0 \boldsymbol{e}_{||}\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\beta_0 p_{||}-E\right]\\ +\Rightarrow 2|\boldsymbol{v}_1| &= \left(\frac{m_2}{m_1}+1\right)|\boldsymbol{v}_4|\\ 
- &= p_{||}\boldsymbol{e}_{||} +p_{T}\boldsymbol{e}_T + \left[\frac{(\gamma_0\beta_0)^2}{\gamma_0+1} p_{||}-\beta_0 \gamma_0 E\right]\boldsymbol{e}_{||}\\ +\Rightarrow |\boldsymbol{v}_4| 
- &= p_{||}\boldsymbol{e}_{||} +p_{T}\boldsymbol{e}_T + \left[\frac{\gamma_0^2-1}{\gamma_0+1} p_{||}-\beta_0 \gamma_0 E\right]\boldsymbol{e}_{||}\\ +&= 2\frac{1}{\frac{m_2}{m_1}+1}|\boldsymbol{v}_1|\\ 
- &= p_{||}\boldsymbol{e}_{||} +p_{T}\boldsymbol{e}_T + \left[(\gamma_0-1) p_{||}-\beta_0 \gamma_0 E\right]\boldsymbol{e}_{||}\\ +&= \frac{2m_1}{m_1+m_2}|\boldsymbol{v}_1|\\ 
- &= \left(\gamma_0 p_{||}-\beta_0 \gamma_0 E\right)\boldsymbol{e}_{||} +p_{T}\boldsymbol{e}_T.\\+\Rightarrow \boldsymbol{v}_4^2 &= \frac{4m_1^2}{(m_1+m_2)^2}\boldsymbol{v}_1^2\\ 
 +\Rightarrow \frac{1}{2}m_4\boldsymbol{v}_4^2 &= \frac{4m_4m_1}{(m_1+m_2)^2}\frac{1}{2}m_1\boldsymbol{v}_1^2\\ 
 +\Rightarrow T_\mathrm{max} &= T_4 = \frac{4m_4m_1}{(m_1+m_2)^2}T_1\\
 \end{align} \end{align}
-Finally,+ 
 +** The second formula for elastic scattering ** 
 + 
 +From the first formula,
 \begin{align} \begin{align}
-\begin{pmatrix} +T_\mathrm{max} 
-E^*\\ +&=\frac{4m_1m_2}{(m_1+m_2)^2}T_1\\ 
-p_{||}^*\\ +&=\frac{2m_2\boldsymbol{p}_1^2}{(m_1+m_2)^2}\\ 
-p_T^*\\ +&=\frac{2m_2m_1^2\boldsymbol{v}_1^2}{(m_1+m_2)^2}\\ 
-\end{pmatrix} +&=\frac{2m_2\boldsymbol{v}_1^2}{(1+m_2/m_1)^2}\\ 
-= +\end{align} 
-\begin{pmatrix} +If $m_2/m_1 \ll 1$, 
-\gamma_0 & -\beta_0\gamma_0 & 0 \\ +\begin{align} 
--\beta_0\gamma_0 & \gamma_0 & 0 \\ +T_\mathrm{max} 
-0 & 0 & 1 +&=\frac{2m_2\boldsymbol{v}_1^2}{(1+m_2/m_1)^2}\\ 
-\end{pmatrix} +&\approx 2m_2\boldsymbol{v}_1^2.\\
-\begin{pmatrix} +
-E\\ +
-p_{||}\\ +
-p_T\\ +
-\end{pmatrix}+
 \end{align} \end{align}
  
 +** Another derivation of the second formula for elastic scattering **
  
 +If $m_2/m_1 \ll 1$, $|\boldsymbol{v}_\mathrm{cm}| \approx |\boldsymbol{v}_1|$.
 +Therefore, from the formula above,
 +\begin{align}
 +T_\mathrm{max}
 +&= 2m_2 \boldsymbol{v}_\mathrm{cm}^2\\
 +&\approx 2m_2\boldsymbol{v}_1^2.\\
 +\end{align}
  
-== Special case ==+** The formula for N-N scattering **
  
-The special case where $\boldsymbol{\beta}_0 = \beta_0 \boldsymbol{k}$, with $\boldsymbol{k}$ the unit vector along the $z$-direction, can be written+For N-N scattering, $m_1 = m_2 = m_\mathrm{N}$. Therefore,  from the formula above, 
 +\begin{align} 
 +T_\mathrm{max} 
 +&=\frac{4m_1m_2}{(m_1+m_2)^2}T_1\\ 
 +&=\frac{4m_\mathrm{N}m_\mathrm{N}}{(m_\mathrm{N}+m_\mathrm{N})^2}T_1\\ 
 +&=\frac{4m_\mathrm{N}^2}{(2m_\mathrm{N})^2}T_1\\ 
 +&=T_1 
 +\end{align}
  
 +==== General memo ====
 +=== Galilean transformation ===
 \begin{align} \begin{align}
-E^*&=\gamma_0E-\beta_0\gamma_0p_z\\ +\boldsymbol{p}^* 
-\boldsymbol{p}^*&=\boldsymbol{p}+\beta_0\boldsymbol{k}\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\beta_0 \boldsymbol{k}\cdot\boldsymbol{p}-E\right]\\ +&=m(\boldsymbol{v}-\boldsymbol{v}_0)\\ 
-                &=\boldsymbol{p}+\beta_0\boldsymbol{k}\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\beta_0 p_z-E\right]\\ +&=\boldsymbol{p}-m\boldsymbol{v}_0
-                &=\boldsymbol{p}+\left[\frac{(\beta_0\gamma_0)^2}{\gamma_0+1}p_z-\beta_0\gamma_0E\right]\boldsymbol{k}\\ +
-                &=\boldsymbol{p}+\left[\frac{\gamma_0^2-1}{\gamma_0+1}p_z-\beta_0\gamma_0E\right]\boldsymbol{k}\\ +
-                &=\boldsymbol{p}+\left[(\gamma_0-1)p_z-\beta_0\gamma_0E\right]\boldsymbol{k}\\ +
-p_x^*\boldsymbol{i}+p_y^*\boldsymbol{j}+p_z^*\boldsymbol{k}&=p_x\boldsymbol{i}+p_y\boldsymbol{j}+p_z\boldsymbol{k}+\left[(\gamma_0-1)p_z-\beta_0\gamma_0E\right]\boldsymbol{k}\\ +
-p_x^*\boldsymbol{i}+p_y^*\boldsymbol{j}+p_z^*\boldsymbol{k}&=p_x\boldsymbol{i}+p_y\boldsymbol{j}+(\gamma_0 p_z-\beta_0\gamma_0E)\boldsymbol{k}\\ +
-p_x^* &= p_x \\ +
-p_y^* &= p_y \\ +
-p_z^* &= \gamma_0 p_z-\beta_0\gamma_0E+
 \end{align} \end{align}
 +
 +So, this equation can be written in another way.
  
 \begin{align} \begin{align}
 \begin{pmatrix} \begin{pmatrix}
-E^*\\ +m\\ 
-p_x^*\\ +\boldsymbol{p}^*
-p_y^*\\ +
-p_z^*+
 \end{pmatrix} \end{pmatrix}
 = =
 \begin{pmatrix} \begin{pmatrix}
-\gamma_0 & 0 & 0 & -\beta_0\gamma_0 \\ +1 & 0 \\ 
-0 & 1 & 0 & 0 \\ +-\boldsymbol{v}_0 & 1\\
-0 & 0 & 1 & 0 \\ +
--\beta_0\gamma_0 & 0 & 0 & \gamma_0+
 \end{pmatrix} \end{pmatrix}
 \begin{pmatrix} \begin{pmatrix}
-E\\ +m\\ 
-p_x\\ +\boldsymbol{p}
-p_y\\ +
-p_z+
 \end{pmatrix} \end{pmatrix}
 \end{align} \end{align}
  
- +=== Velocity of the center of mass $\boldsymbol{v}_\mathrm{cm}$ ===
-=== Velocity of the center of mass $\boldsymbol{\beta}_{\rm cm}$ ===+
  
 In the center of mass frame, the sum of the momenta of the particles ($\boldsymbol{p}_1^* + \boldsymbol{p}_2^*$) should be $0$. In the center of mass frame, the sum of the momenta of the particles ($\boldsymbol{p}_1^* + \boldsymbol{p}_2^*$) should be $0$.
-Therefore, $\boldsymbol{\beta}_{\rm cm}$ can be obtained by solving the following equation for $\boldsymbol{\beta}_0$.+Therefore, $\boldsymbol{v}_{\rm cm}$ can be obtained by solving the following equation for $\boldsymbol{v}_0$.
 \begin{align} \begin{align}
 0 &= \boldsymbol{p}_1^* + \boldsymbol{p}_2^*\\ 0 &= \boldsymbol{p}_1^* + \boldsymbol{p}_2^*\\
-  &= \boldsymbol{p}_1+\boldsymbol{\beta}_0\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\boldsymbol{\beta}_0\cdot\boldsymbol{p}_1-E_1\right] + \boldsymbol{p}_2+\boldsymbol{\beta}_0\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\boldsymbol{\beta}_0\cdot\boldsymbol{p}_2-E_2\right]\\+  &= \boldsymbol{p}_1-m_1\boldsymbol{v}_0 + \boldsymbol{p}_2-m_2\boldsymbol{v}_0\\
 \end{align} \end{align}
-The equation can be rearranged as below.+By solving the equation for $\boldsymbol{v}_0$,
 \begin{align} \begin{align}
-\boldsymbol{p}_1 + \boldsymbol{p}_2 = -\boldsymbol{\beta}_0\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1+\boldsymbol{p}_2)-(E_1+E_2)\right]+\boldsymbol{v}_0 = \frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{m_1+m_2}.
 \end{align} \end{align}
-This equation means that the direction of $\boldsymbol{\beta}_0$ is the same as that of $(\boldsymbol{p}_1 + \boldsymbol{p}_2)$. (Note that $\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1+\boldsymbol{p}_2)$ is a scalar term and has no direction.) Therefore, the unit vector $\widehat{\boldsymbol{\beta}_0}$ along the direction of $\boldsymbol{\beta}_0$ is +Therefore, the $\boldsymbol{v}_\mathrm{cm}$ is
 \begin{align} \begin{align}
-\widehat{\boldsymbol{\beta}_0} = \frac{\boldsymbol{\beta}_0}{\beta_0} = \frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}. +\boldsymbol{v}_\mathrm{cm}  
-\end{align} +&= \frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{m_1+m_2}\\ 
-The magnitude $\beta_0=|\boldsymbol{\beta}_0|$ of the vector $\boldsymbol{\beta}_0$ can be calculated by making inner product of $\boldsymbol{\beta}_0$ and $(\boldsymbol{p}_1 + \boldsymbol{p}_2)$ as below. +&= \frac{m_1\boldsymbol{v}_1 + m_2\boldsymbol{v}_2}{m_1+m_2}
-\begin{align} +
-\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2) +
-&= -\boldsymbol{\beta}_0\cdot\boldsymbol{\beta}_0\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1+\boldsymbol{p}_2)-(E_1+E_2)\right]\\ +
-&= -\beta_0^2\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1+\boldsymbol{p}_2)-(E_1+E_2)\right]\\ +
-&= -\frac{(\beta_0\gamma_0)^2}{\gamma_0+1}\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1+\boldsymbol{p}_2)+\beta_0^2\gamma_0(E_1+E_2)\\ +
-\beta_0^2\gamma_0(E_1+E_2) +
-&= \left[1+\frac{(\beta_0\gamma_0)^2}{\gamma_0+1}\right]\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2)\\ +
-&= \left[1+\frac{\gamma_0^2-1}{\gamma_0+1}\right]\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2)\\ +
-&= \left[1+(\gamma_0-1)\right]\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2)\\ +
-&= \gamma_0\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2)\\ +
-\beta_0^2(E_1+E_2) +
-&= \boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2)\\ +
-\beta_0(E_1+E_2) +
-&= \left(\frac{\boldsymbol{\beta}_0}{\beta_0}\right)\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2)\\ +
-&= \left(\frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\right)\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2)\\ +
-&= \frac{(\boldsymbol{p}_1 + \boldsymbol{p}_2)\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2)}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\\ +
-&= \frac{|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\\ +
-&=|\boldsymbol{p}_1 + \boldsymbol{p}_2|\\ +
-\beta_0&=\frac{|\boldsymbol{p}_1 + \boldsymbol{p}_2|}{E_1+E_2}. +
-\end{align} +
-Therefore, +
-\begin{align} +
-\widehat{\boldsymbol{\beta}_{\rm cm}} = \frac{\boldsymbol{\beta}_{\rm cm}}{\beta_{\rm cm}} = \frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|},\\ +
-\beta_{\rm cm} = \frac{|\boldsymbol{p}_1 + \boldsymbol{p}_2|}{E_1+E_2}. +
-\end{align} +
-As a result, +
-\begin{align} +
-\boldsymbol{\beta}_{\rm cm} = \beta_{\rm cm}\widehat{\boldsymbol{\beta}_{\rm cm}} = \frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{E_1+E_2}. +
-\end{align} +
-In other words, +
-\begin{align} +
-\boldsymbol{\beta}_{\rm cm} = \frac{E_1\boldsymbol{\beta}_1 + E_2\boldsymbol{\beta}_2}{E_1+E_2}.+
 \end{align} \end{align}
  
-N.B. In Galilean transformation (non-relativistic limit), 
-\begin{align} 
-\boldsymbol{\beta}_{\rm cm} = \frac{m_1\boldsymbol{\beta}_1 + m_2\boldsymbol{\beta}_2}{m_1+m_2}. 
-\end{align} 
- 
-==== The total center of mass energy $W$, $\gamma_{\rm cm}$ and $\boldsymbol{\beta}_{\rm cm}\gamma_{\rm cm}$==== 
-\begin{align} 
-W &= \sqrt{(E_1+E_2)^2-(\boldsymbol{p}_1+\boldsymbol{p}_2)^2}\\ 
-\gamma_{\rm cm} 
-&= \frac{1}{\sqrt{1-\beta_{\rm cm}^2}}\\ 
-&= \frac{1}{\sqrt{1-\left(\frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{E_1+E_2}\right)^2}}\\ 
-&= \frac{E_1+E_2}{\sqrt{(E_1+E_2)^2-(\boldsymbol{p}_1 + \boldsymbol{p}_2)^2}}\\ 
-&= \frac{E_1+E_2}{W}\\ 
-\boldsymbol{\beta}_{\rm cm}\gamma_{\rm cm} 
-&= \frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{E_1+E_2} \frac{E_1+E_2}{W} \\ 
-&= \frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{W} \\ 
-\beta_{\rm cm}\gamma_{\rm cm} 
-&= \frac{|\boldsymbol{p}_1 + \boldsymbol{p}_2|}{W} 
-\end{align} 
- 
-=== The momentum $\boldsymbol{p}_1^*$ in the center of mass frame === 
-\begin{align} 
-\boldsymbol{p}_1^* 
- &= \boldsymbol{p}_1+\boldsymbol{\beta}_{\rm cm}\gamma_{\rm cm}\left[\frac{\gamma_{\rm cm}}{\gamma_{\rm cm}+1}\boldsymbol{\beta}_{\rm cm}\cdot\boldsymbol{p}_1-E_1\right]\\ 
- &= \boldsymbol{p}_1+\beta_{\rm cm}\widehat{\boldsymbol{\beta}_{\rm cm}}\gamma_{\rm cm}\left[\frac{\gamma_{\rm cm}}{\gamma_{\rm cm}+1}\beta_{\rm cm}\widehat{\boldsymbol{\beta}_{\rm cm}}\cdot\boldsymbol{p}_1-E_1\right]\\ 
- &= \boldsymbol{p}_1+\widehat{\boldsymbol{\beta}_{\rm cm}}\left[\frac{(\beta_{\rm cm}\gamma_{\rm cm})^2}{\gamma_{\rm cm}+1}\widehat{\boldsymbol{\beta}_{\rm cm}}\cdot\boldsymbol{p}_1-\beta_{\rm cm}\gamma_{\rm cm}E_1\right]\\ 
- &= \boldsymbol{p}_1+\widehat{\boldsymbol{\beta}_{\rm cm}}\left[(\gamma_{\rm cm}-1)\widehat{\boldsymbol{\beta}_{\rm cm}}\cdot\boldsymbol{p}_1-\beta_{\rm cm}\gamma_{\rm cm}E_1\right]\\ 
- &= \boldsymbol{p}_1+\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\left[\left(\frac{E_1+E_2}{W}-1\right)\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\cdot\boldsymbol{p}_1-\frac{|\boldsymbol{p}_1+\boldsymbol{p}_2|}{W}E_1\right]\\ 
- &= \boldsymbol{p}_1+\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\left[\frac{E_1+E_2}{W}\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\cdot\boldsymbol{p}_1-\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\cdot\boldsymbol{p}_1-\frac{|\boldsymbol{p}_1+\boldsymbol{p}_2|}{W}E_1\right]\\ 
- &= \boldsymbol{p}_1+\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\left[\frac{E_1+E_2}{W}\frac{\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}-\frac{\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}-\frac{|\boldsymbol{p}_1+\boldsymbol{p}_2|}{W}E_1\right]\\ 
- &= \boldsymbol{p}_1+\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\left[\frac{(E_1+E_2)(\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2)-E_1|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|}-\frac{\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\right]\\ 
- &= \boldsymbol{p}_1+\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\left[\frac{E_1(\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2)+E_2(\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2)-E_1(\boldsymbol{p}_1^2+\boldsymbol{p}_2^2+2\boldsymbol{p}_1\cdot\boldsymbol{p}_2)}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|}-\frac{\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\right]\\ 
- &= \boldsymbol{p}_1+\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\left[\frac{\boldsymbol{p}_1^2E_2-\boldsymbol{p}_2^2E_1+(E_2-E_1)\boldsymbol{p}_1\cdot\boldsymbol{p}_2}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|}-\frac{\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\right]\\ 
- &=  \frac{\left[\boldsymbol{p}_1^2E_2-\boldsymbol{p}_2^2E_1+(E_2-E_1)\boldsymbol{p}_1\cdot\boldsymbol{p}_2\right](\boldsymbol{p}_1+\boldsymbol{p}_2)}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\boldsymbol{p}_1-\frac{(\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2)(\boldsymbol{p}_1+\boldsymbol{p}_2)}{|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ 
- &=  \frac{\left[\boldsymbol{p}_1^2E_2-\boldsymbol{p}_2^2E_1+(E_2-E_1)\boldsymbol{p}_1\cdot\boldsymbol{p}_2\right](\boldsymbol{p}_1+\boldsymbol{p}_2)}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\frac{|\boldsymbol{p}_1+\boldsymbol{p}_2|^2\boldsymbol{p}_1-(\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2)\boldsymbol{p}_1-(\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2)\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ 
- &=  \frac{\left[\boldsymbol{p}_1^2E_2-\boldsymbol{p}_2^2E_1+(E_2-E_1)\boldsymbol{p}_1\cdot\boldsymbol{p}_2\right](\boldsymbol{p}_1+\boldsymbol{p}_2)}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\frac{(\boldsymbol{p}_1^2+\boldsymbol{p}_2^2+2\boldsymbol{p}_1\cdot\boldsymbol{p}_2)\boldsymbol{p}_1-(\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2)\boldsymbol{p}_1-(\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2)\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ 
- &= \frac{\left[\boldsymbol{p}_1^2E_2-\boldsymbol{p}_2^2E_1+(E_2-E_1)\boldsymbol{p}_1\cdot\boldsymbol{p}_2\right](\boldsymbol{p}_1+\boldsymbol{p}_2)}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\frac{\boldsymbol{p}_2^2\boldsymbol{p}_1-\boldsymbol{p}_1^2\boldsymbol{p}_2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2(\boldsymbol{p}_1-\boldsymbol{p}_2)}{|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ 
- &= \frac{\left[\boldsymbol{p}_1^2E_2-\boldsymbol{p}_2^2E_1+(E_2-E_1)\cdot\frac{1}{2}\left(|\boldsymbol{p}_1+\boldsymbol{p}_2|^2-\boldsymbol{p}_1^2-\boldsymbol{p}_2^2\right)\right](\boldsymbol{p}_1+\boldsymbol{p}_2)}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\frac{\boldsymbol{p}_2^2\boldsymbol{p}_1-\boldsymbol{p}_1^2\boldsymbol{p}_2+\frac{1}{2}\left(|\boldsymbol{p}_1+\boldsymbol{p}_2|^2-\boldsymbol{p}_1^2-\boldsymbol{p}_2^2\right)(\boldsymbol{p}_1-\boldsymbol{p}_2)}{|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ 
- &= \frac{\left[\boldsymbol{p}_1^2E_2-\boldsymbol{p}_2^2E_1-\frac{1}{2}(E_2-E_1)(\boldsymbol{p}_1^2+\boldsymbol{p}_2^2)+\frac{1}{2}(E_2-E_1)|\boldsymbol{p}_1+\boldsymbol{p}_2|^2\right](\boldsymbol{p}_1+\boldsymbol{p}_2)}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\frac{\boldsymbol{p}_2^2\boldsymbol{p}_1-\boldsymbol{p}_1^2\boldsymbol{p}_2-\frac{1}{2}(\boldsymbol{p}_1^2+\boldsymbol{p}_2^2)(\boldsymbol{p}_1-\boldsymbol{p}_2)+\frac{1}{2}|\boldsymbol{p}_1+\boldsymbol{p}_2|^2(\boldsymbol{p}_1-\boldsymbol{p}_2)}{|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ 
- &= \frac{\left[\frac{1}{2}\boldsymbol{p}_1^2E_2-\frac{1}{2}\boldsymbol{p}_2^2E_1+\frac{1}{2}\boldsymbol{p}_1^2E_1-\frac{1}{2}\boldsymbol{p}_2^2E_2+\frac{1}{2}(E_2-E_1)|\boldsymbol{p}_1+\boldsymbol{p}_2|^2\right](\boldsymbol{p}_1+\boldsymbol{p}_2)}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\frac{\frac{1}{2}\boldsymbol{p}_2^2\boldsymbol{p}_1-\frac{1}{2}\boldsymbol{p}_1^2\boldsymbol{p}_2-\frac{1}{2}\boldsymbol{p}_1^2\boldsymbol{p}_1+\frac{1}{2}\boldsymbol{p}_2^2\boldsymbol{p}_2+\frac{1}{2}|\boldsymbol{p}_1+\boldsymbol{p}_2|^2(\boldsymbol{p}_1-\boldsymbol{p}_2)}{|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ 
- &= \frac{\left[(E_1+E_2)(\boldsymbol{p}_1^2-\boldsymbol{p}_2^2)+(E_2-E_1)|\boldsymbol{p}_1+\boldsymbol{p}_2|^2\right](\boldsymbol{p}_1+\boldsymbol{p}_2)}{2W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\frac{(\boldsymbol{p}_2^2-\boldsymbol{p}_1^2)(\boldsymbol{p}_1+\boldsymbol{p}_2)+|\boldsymbol{p}_1+\boldsymbol{p}_2|^2(\boldsymbol{p}_1-\boldsymbol{p}_2)}{2|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ 
- &= \frac{\left[(E_1+E_2)(\boldsymbol{p}_1^2-\boldsymbol{p}_2^2)+(E_2-E_1)|\boldsymbol{p}_1+\boldsymbol{p}_2|^2\right](\boldsymbol{p}_1+\boldsymbol{p}_2)+W(\boldsymbol{p}_2^2-\boldsymbol{p}_1^2)(\boldsymbol{p}_1+\boldsymbol{p}_2)}{2W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\frac{1}{2}(\boldsymbol{p}_1-\boldsymbol{p}_2)\\ 
- &= \frac{(E_1+E_2-W)(\boldsymbol{p}_1^2-\boldsymbol{p}_2^2)+(E_2-E_1)|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}{2W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}(\boldsymbol{p}_1+\boldsymbol{p}_2)+\frac{1}{2}(\boldsymbol{p}_1-\boldsymbol{p}_2)\\ 
- &= \frac{(E_1+E_2-W)(\boldsymbol{p}_1^2-\boldsymbol{p}_2^2)(\boldsymbol{p}_1+\boldsymbol{p}_2)+(E_2-E_1)|\boldsymbol{p}_1+\boldsymbol{p}_2|^2(\boldsymbol{p}_1+\boldsymbol{p}_2)+W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2(\boldsymbol{p}_1-\boldsymbol{p}_2)}{2W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ 
- &= \frac{\left[(E_1+E_2-W)(\boldsymbol{p}_1^2-\boldsymbol{p}_2^2)+(E_2-E_1+W)|\boldsymbol{p}_1+\boldsymbol{p}_2|^2\right]\boldsymbol{p}_1+\left[(E_1+E_2-W)(\boldsymbol{p}_1^2-\boldsymbol{p}_2^2)+(E_2-E_1-W)|\boldsymbol{p}_1+\boldsymbol{p}_2|^2\right]\boldsymbol{p}_2}{2W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ 
- &= \frac{\left[(E_1+E_2-W)(E_1^2-m_1^2-E_2^2+m_2^2)+(E_2-E_1+W)\left[(E_1+E_2)^2-W^2\right]\right]\boldsymbol{p}_1+\left[(E_1+E_2-W)(E_1^2-m_1^2-E_2^2+m_2^2)+(E_2-E_1-W)\left[(E_1+E_2)^2-W^2\right]\right]\boldsymbol{p}_2}{2W\left[(E_1+E_2)^2-W^2\right]}\\ 
- &= \frac{\left[(E_1+E_2-W)(E_1^2-m_1^2-E_2^2+m_2^2)+(E_2-E_1+W)(E_1+E_2+W)(E_1+E_2-W)\right]\boldsymbol{p}_1+\left[(E_1+E_2-W)(E_1^2-m_1^2-E_2^2+m_2^2)+(E_2-E_1-W)(E_1+E_2+W)(E_1+E_2-W)\right]\boldsymbol{p}_2}{2W(E_1+E_2+W)(E_1+E_2-W)}\\ 
- &= \frac{\left[(E_1^2-m_1^2-E_2^2+m_2^2)+(E_2-E_1+W)(E_1+E_2+W)\right]\boldsymbol{p}_1+\left[(E_1^2-m_1^2-E_2^2+m_2^2)+(E_2-E_1-W)(E_1+E_2+W)\right]\boldsymbol{p}_2}{2W(E_1+E_2+W)}\\ 
- &= \frac{\left(m_2^2-m_1^2+2E_2W+W^2\right)\boldsymbol{p}_1+\left(m_2^2-m_1^2-2E_1W-W^2\right)\boldsymbol{p}_2}{2W(E_1+E_2+W)}\\ 
-\end{align} 
  
research/memos/kinematics/non-relativistic_kinematics.1506663173.txt.gz · 最終更新: 2017/09/29 14:32 by kobayash
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