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research:memos:kinematics:non-relativistic_kinematics [2017/10/01 18:16] – [Formulae for Non-relativistic Kinematics] 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= 2m_\mathrm{N}+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}
行 47: 行 51:
 \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 54: 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 272: 行 275:
 \cos\theta_3 \cos\theta_3
 &= \frac{1}{\sqrt{1+\left[\frac{\sin\theta_3^*}{\delta_{23}^*+\cos\theta_3^*}\right]^2}}\\ &= \frac{1}{\sqrt{1+\left[\frac{\sin\theta_3^*}{\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}}\\ +&= \frac{\delta_{23}^*+\cos\theta_3^*}{\sqrt{\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}
-&= \frac{\delta_{23}^*+\cos\theta_3^*}{\sqrt{\sin^2\theta_3^*+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*+\cos^2\theta_3^*}}\\ +
-&= \frac{\delta_{23}^*+\cos\theta_3^*}{\sqrt{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*}}+
 \end{align} \end{align}
  
行 358: 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}^*\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}}.+\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 392: 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{\delta_{23}^*+\cos\theta_3^*}{\sqrt{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*}}\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(\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[1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*\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[\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[1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*\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[\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[1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*\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[\left[1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*\right]-\delta_{23}^{*}\left(\delta_{23}^*+\cos\theta_3^*\right)\right]\right/\left[1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*\right]^{3/2}\\ +&= \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[1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*-\delta_{23}^{*2}-\delta_{23}^{*}\cos\theta_3^*\right]\right/\left[1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*\right]^{3/2}\\ +&= \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}\\ 
-&= \frac{1+\delta_{23}^{*}\cos\theta_3^*}{\left[1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*\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}\\ 
 +&= \frac{1+\delta_{23}^{*}\cos\theta_3^*}{\left[\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2\right]^{3/2}}
 \end{align} \end{align}
  
行 406: 行 408:
 From Quantity 7., From Quantity 7.,
 \begin{align} \begin{align}
-\cos\theta_3 &= \frac{\delta_{23}^*+\cos\theta_3^*}{\sqrt{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*}}.+\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{\delta_{23}^*+\cos\theta_3^*}{\sqrt{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*}}\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{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*-\left(\delta_{23}^*+\cos\theta_3^*\right)^2}{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*}}\\ +&= \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{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)}{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*}}\\ +&= \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}}\\ 
-&= \sqrt{\frac{1-\cos^2\theta_3^*}{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*}}\\ +&= \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^*}{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*}}\\ +&= \sqrt{\frac{\sin^2\theta_3^*}{\sin^2\theta_3^*+\left(\delta_{23}^*+\cos\theta_3^*\right)^2}}\\ 
-&= \frac{\sin\theta_3^*}{\sqrt{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*}}\\+&= \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{1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*}}+&= \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{1+\delta_{23}^*\cos\theta_3^*}{\left[1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*\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^*}(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 433: 行 436:
  
 \begin{align} \begin{align}
-\frac{d(\cos\theta_3)}{d(\cos\theta_3^*)}+\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{1+\delta_{23}^{*}\cos\theta_3^*}{\left[1+\delta_{23}^{*2}+2\delta_{23}^*\cos\theta_3^*\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+2\cos\theta_3^*\right]^{3/2}}\\
行 516: 行 520:
 === 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}
行 537: 行 541:
 \end{align} \end{align}
  
-Therefore, $|\boldsymbol{v}_4|$ becomes maximum at $\theta_4^* = \theta_\mathrm{cm}=\pi$. Because of $\cos\theta_4^*=-1$, the maximum value is+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}
-|\boldsymbol{v}_4| &= |\boldsymbol{v}_4^*|+v_{\rm cm}.+T_1 &= T_3 + T_4\\ 
 +\boldsymbol{p}_1 &= \boldsymbol{p}_3 + \boldsymbol{p}_4.
 \end{align} \end{align}
  
-In this case, from the law of conservation of energy and momentum, +In general, $T = \frac{\boldsymbol{p}^2}{2m}$. Therefore, the first equation is written as
 \begin{align} \begin{align}
-T_1 &= T_3 + T_4\\ +\frac{\boldsymbol{p}_1^2}{2m_1} &= \frac{\boldsymbol{p}_3^2}{2m_3} + \frac{\boldsymbol{p}_4^2}{2m_4}\\ 
-|\boldsymbol{p}_1| &= |\boldsymbol{p}_3| + |\boldsymbol{p}_4|+\Rightarrow \frac{\boldsymbol{p}_1^2}{m_1} &= \frac{\boldsymbol{p}_3^2}{m_3} + \frac{\boldsymbol{p}_4^2}{m_4}
 \end{align} \end{align}
  
-In general, $\sqrt{2mT} = |\boldsymbol{p}|$. Therefore, the second equation is written as+By substituting $\boldsymbol{p}_3 = \boldsymbol{p}_1 - \boldsymbol{p}_4$ into this formula,  
 \begin{align} \begin{align}
-\sqrt{2m_1T_1} &= \sqrt{2m_3T_3} + \sqrt{2m_4T_4}\\ +\frac{\boldsymbol{p}_1^2}{m_1} 
-\Rightarrow \sqrt{m_1T_1} &= \sqrt{m_3T_3} + \sqrt{m_4T_4} +&= \frac{(\boldsymbol{p}_1 - \boldsymbol{p}_4)^2}{m_3} + \frac{\boldsymbol{p}_4^2}{m_4}\\ 
 +&= \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}\\ 
 +&= \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}.
 \end{align} \end{align}
  
-By substituting $T_3=T_1-T_4$ into this formula, +Therefore, by using $\boldsymbol{p}^2=|\boldsymbol{p}|^2$,
  
 \begin{align} \begin{align}
-\sqrt{m_1T_1} &= \sqrt{m_3(T_1-T_4)} + \sqrt{m_4T_4}\\ +\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\\
-\Rightarrow \sqrt{m_1T_1} - \sqrt{m_4T_4} &= \sqrt{m_3(T_1-T_4)}.+
 \end{align} \end{align}
  
-By taking the square for each side,+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}
-m_1T_1 -2\sqrt{m_1m_4T_1T_4} + m_4T_4  &= m_3(T_1-T_4)\\ +|\boldsymbol{p}_4| 
-\Rightarrow (m_1-m_3)T_1 + (m_3+m_4)T_4 & = 2\sqrt{m_1m_4T_1T_4}.+&=\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}}\\ 
 +&=\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|\\ 
 +&=\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|\\ 
 +&=\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 taking the square for each side again,+The direction of $\boldsymbol{p}_4$ is the same as $\boldsymbol{p}_1$, then 
 +\begin{align} 
 +\boldsymbol{p}_4 
 +&=\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}
  
 +From $\boldsymbol{p}_3=\boldsymbol{p}_1-\boldsymbol{p}_4$,
 \begin{align} \begin{align}
-(m_1-m_3)^2T_1^2 +2(m_1-m_3)(m_3+m_4)T_1T_4 + (m_3+m_4)^2T_4^2 & = 4m_1m_4T_1T_4\\ +\boldsymbol{p}_3 
-\Rightarrow (m_3+m_4)^2T_4^2 + 2\left[(m_1-m_3)(m_3+m_4)-2m_1m_4\right]T_1T_4 + (m_1-m_3)^2T_1^2 & = 0.+&= \boldsymbol{p}_1-\boldsymbol{p}_4\\ 
 +&= \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{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}
  
-In general, the solution of the equation $ax^2+2bx+c=0$ is $x=\frac{-b\pm\sqrt{b^2-ac}}{a}$. Therefore,+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_4 T_4
-&= \frac{-\left[(m_1-m_3)(m_3+m_4)-2m_1m_4\right]T_1\pm\sqrt{\left[(m_1-m_3)(m_3+m_4)-2m_1m_4\right]^2T_1^2-(m_1-m_3)^2(m_3+m_4)^2T_1^2}}{(m_3+m_4)^2}\\ +&=\frac{|\boldsymbol{p}_4|^2}{2m_4}\\ 
-&= \frac{-\left[(m_1-m_3)(m_3+m_4)-2m_1m_4\right]\pm\sqrt{\left[(m_1-m_3)(m_3+m_4)-2m_1m_4\right]^2-(m_1-m_3)^2(m_3+m_4)^2}}{(m_3+m_4)^2}T_1\\ +&=\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{-\left[(m_1-m_3)(m_3+m_4)-2m_1m_4\right]\pm\sqrt{(m_1-m_3)^2(m_3+m_4)^2-4m_1m_4(m_1-m_3)(m_3+m_4)+4m_1^2m_4^2-(m_1-m_3)^2(m_3+m_4)^2}}{(m_3+m_4)^2}T_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}\frac{|\boldsymbol{p}_1|^2}{2m_1}\\ 
-&= \frac{-\left[(m_1-m_3)(m_3+m_4)-2m_1m_4\right]\pm\sqrt{-4m_1m_4(m_1-m_3)(m_3+m_4)+4m_1^2m_4^2}}{(m_3+m_4)^2}T_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[(m_1-m_3)(m_3+m_4)-2m_1m_4\right]\pm2\sqrt{m_1m_4}\sqrt{m_1m_4-(m_1-m_3)(m_3+m_4)}}{(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[(m_1-m_3)(m_3+m_4)-2m_1m_4\right]\pm2\sqrt{m_1m_4}\sqrt{m_1m_4-[m_1m_4+(m_1-m_4)m_3-m_3^2]}}{(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.
-&= \frac{-\left[(m_1-m_3)(m_3+m_4)-2m_1m_4\right]\pm2\sqrt{m_1m_4}\sqrt{m_3^2-(m_1-m_4)m_3}}{(m_3+m_4)^2}T_1\\ +
-&= \frac{-\left[(m_1-m_3)(m_3+m_4)-2m_1m_4\right]\pm2\sqrt{m_1m_3m_4}\sqrt{m_3-(m_1-m_4)}}{(m_3+m_4)^2}T_1\\ +
-&= \frac{-\left[(m_1-m_3)(m_3+m_4)-2m_1m_4\right]\pm2\sqrt{m_1m_3m_4}\sqrt{m_3+m_4-m_1}}{(m_3+m_4)^2}T_1\\+
 \end{align} \end{align}
  
 +As a result,
  
 +\begin{align}
 +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.
 +\end{align}
  
 +By the way, from $T_3=T_1-T_4$, $T_3$ is  
  
- 
-From this formula, 
 \begin{align} \begin{align}
-T_\mathrm{max} +T_3 
-&= E_\mathrm{4max} - m_4\\ +&= T_1-T_4\\ 
-&= \gamma_\mathrm{cm} (E_4^* + \beta_\mathrm{cm}|\boldsymbol{p}_4^*|) - m_4.+&= 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}
  
-For the elastic scattering, $m_4 = m_2$, $|\boldsymbol{p}_4^*| = |\boldsymbol{p}_2^*|$, and $|E_4^*| = |E_2^*|$. Then,+** The first formula for elastic scattering ** 
 + 
 +For elastic scattering, $m_3 = m_1$ and $m_4 = m_2$. Therefore,
 \begin{align} \begin{align}
-T_\mathrm{max} +\boldsymbol{p}_4&=\frac{2m_2}{m_1+m_2}\boldsymbol{p}_1\\ 
-&= \gamma_\mathrm{cm} (E_2^* + \beta_\mathrm{cm}|\boldsymbol{p}_2^*|) - m_2\\ +\boldsymbol{p}_3&=\frac{m_1-m_2}{m_1+m_2}\boldsymbol{p}_1\\ 
-&= \gamma_\mathrm{cm} [m_2\gamma_2^* + \beta_\mathrm{cm}(m_2|\boldsymbol{\beta}_2^*|\gamma_2^*)] - m_2.+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}
-By using $|\boldsymbol{\beta}_2^*| = \beta_\mathrm{cm}$ and $\gamma_2^* = \gamma_\mathrm{cm}$,+As a result, 
 \begin{align} \begin{align}
-T_\mathrm{max} +T_\mathrm{max}=T_4=\frac{4m_1m_2}{(m_1+m_2)^2}T_1
-&= \gamma_\mathrm{cm} [m_2\gamma_\mathrm{cm} + \beta_\mathrm{cm}(m_2\beta_\mathrm{cm}\gamma_\mathrm{cm})] - m_2\\ +
-&= 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,+ 
 + 
 + 
 +** 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}
-T_\mathrm{max} +\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_2\gamma_\mathrm{cm}^2 \left(1+\frac{\gamma_\mathrm{cm}^2-1}{\gamma_\mathrm{cm}^2}\right) - m_2\\ +m_1\boldsymbol{v}_1 = m_1\boldsymbol{v}_3 + m_2\boldsymbol{v}_4\\
-&= m_2 (\gamma_\mathrm{cm}^2+\gamma_\mathrm{cm}^2-1) - m_2\\ +
-&= 2m_2 (\gamma_\mathrm{cm}^2-1)\\ +
-&= 2m_2 \boldsymbol{\beta}_\mathrm{cm}^2\gamma_\mathrm{cm}^2.\\+
 \end{align} \end{align}
-By using Quantity 4. and 5.,+ 
 +From the second equation,
 \begin{align} \begin{align}
-T_\mathrm{max} +\boldsymbol{v}_3 = \frac{m_1\boldsymbol{v}_1 - m_2\boldsymbol{v}_4}{m_1}.
-&= 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+By substituting this formula into the formula of the law of conservation of energy, 
 \begin{align} \begin{align}
-W +\frac{1}{2}m_1\boldsymbol{v}_1^2 
-&= \sqrt{m_1^2+m_2^2+2m_2E_1}\\ +&= \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\\ 
-&= \sqrt{m_1^2+m_2^2+2m_2m_1\gamma_1}\\ +&= \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\\ 
-&= \sqrt{m_1m_2\left(\frac{m_1}{m_2}+\frac{m_2}{m_1}+2\gamma_1\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 \\ 
-&= \sqrt{m_1m_2\left(k_{12}+k_{21}+2\gamma_1\right)}.\\ +\Rightarrow 0 &= \frac{m_2|\boldsymbol{v}_4| - 2m_1|\boldsymbol{v}_1|}{m_1} + |\boldsymbol{v}_4| \\ 
-\end{align} +\Rightarrow 2|\boldsymbol{v}_1| &= \left(\frac{m_2}{m_1}+1\right)|\boldsymbol{v}_4|\\ 
-By substituting this formula into the former formula, +\Rightarrow |\boldsymbol{v}_4| 
-\begin{align} +&= 2\frac{1}{\frac{m_2}{m_1}+1}|\boldsymbol{v}_1|\\ 
-T_\mathrm{max} +&= \frac{2m_1}{m_1+m_2}|\boldsymbol{v}_1|\\ 
-&= 2m_2  \frac{\boldsymbol{p}_1^2}{m_1m_2\left(k_{12}+k_{21}+2\gamma_1\right)}\\ +\Rightarrow \boldsymbol{v}_4^2 &= \frac{4m_1^2}{(m_1+m_2)^2}\boldsymbol{v}_1^2\\ 
-&= \frac{2\boldsymbol{p}_1^2}{m_1(k_{12}+k_{21}+2\gamma_1)}.\\+\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}
  
-** The second formula of the General Formulae **+** The second formula for elastic scattering **
  
 From the first formula, From the first formula,
 \begin{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\\ 
-&= \frac{2m_1^2\boldsymbol{\beta}_1^2\gamma_1^2}{m_1(k_{12}+k_{21}+2\gamma_1)}\\ +&=\frac{2m_2\boldsymbol{p}_1^2}{(m_1+m_2)^2}\\ 
-&= \frac{2m_1\boldsymbol{\beta}_1^2\gamma_1^2}{k_{12}+k_{21}+2\gamma_1}.\\+&=\frac{2m_2m_1^2\boldsymbol{v}_1^2}{(m_1+m_2)^2}\\ 
 +&=\frac{2m_2\boldsymbol{v}_1^2}{(1+m_2/m_1)^2}\\
 \end{align} \end{align}
-By using $m_1=m_2/k_{21}$,+If $m_2/m_1 \ll 1$,
 \begin{align} \begin{align}
 T_\mathrm{max} T_\mathrm{max}
-&= \frac{2m_2\boldsymbol{\beta}_1^2\gamma_1^2}{k_{21}(k_{12}+k_{21}+2\gamma_1)}\\ +&=\frac{2m_2\boldsymbol{v}_1^2}{(1+m_2/m_1)^2}\\ 
-&= \frac{2m_2\boldsymbol{\beta}_1^2\gamma_1^2}{1+k_{21}^2+2\gamma_1k_{21}}.\\+&\approx 2m_2\boldsymbol{v}_1^2.\\
 \end{align} \end{align}
-If $k_{21} = m2/m1 \ll 1$,+ 
 +** 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} \begin{align}
 T_\mathrm{max} T_\mathrm{max}
-&= \frac{2m_2\boldsymbol{\beta}_1^2\gamma_1^2}{1+k_{21}^2+2\gamma_1k_{21}}\\ +&= 2m_2 \boldsymbol{v}_\mathrm{cm}^2\\ 
-&\approx 2m_2\boldsymbol{\beta}_1^2\gamma_1^2.\\+&\approx 2m_2\boldsymbol{v}_1^2.\\
 \end{align} \end{align}
  
-** Another derivation of the second formula of the General Formulae **+** The formula for N-N scattering **
  
-If $m2/m1 \ll 1$, $|\boldsymbol{\beta}_\mathrm{cm}| \approx |\boldsymbol{\beta}_1|$ and $\gamma_\mathrm{cm} \approx \gamma_1$. +For N-N scattering, $m_1 = m_2 = m_\mathrm{N}$. Therefore,  from the formula above,
-Therefore, from the formula above,+
 \begin{align} \begin{align}
 T_\mathrm{max} T_\mathrm{max}
-&= 2m_2 \boldsymbol{\beta}_\mathrm{cm}^2\gamma_\mathrm{cm}^2\\ +&=\frac{4m_1m_2}{(m_1+m_2)^2}T_1\\ 
-&\approx 2m_2\boldsymbol{\beta}_1^2\gamma_1^2.\\+&=\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} \end{align}
- 
  
 ==== General memo ==== ==== General memo ====
research/memos/kinematics/non-relativistic_kinematics.1506849401.txt.gz · 最終更新: 2017/10/01 18:16 (外部編集)
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