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--- research:memos:kinematics:non-relativistic_kinematics [2017/09/29 23:11] – [The total center of mass energy $W$, $\gamma_{\rm cm}$ and $\boldsymbol{\beta}_{\rm cm}\gamma_{\rm cm}$] kobayash
+++ research: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|}}
-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
@@ 行 14: / 行 14: @@
 \delta_{ij} = |\boldsymbol{v}_i|/|\boldsymbol{v}_j|,
 \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)**|
@@ 行 20: / 行 24: @@
 W
 &= m_1 + m_2\\
-&= m_3 + m_4
+&= m_3 + m_4?
 \end{align}| Same as the General formula |\begin{align}
-W= 2m_\mathrm{N}
+W= 2m_\mathrm{N}?
 \end{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}
-|\boldsymbol{p}_1'| = \frac{1}{2}\sqrt{W^2-4m_\mathrm{N}^2}
+\boldsymbol{p}_1' = \frac{\boldsymbol{p}_1}{2}
 \end{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}
 |\boldsymbol{p}_3^{*}| = |\boldsymbol{p}_1^{*}|
 \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}|
 | 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 |
 | 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 |
 | 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{otherwise\ } \theta_{3\mathrm{max}} = 180^\circ
@@ 行 50: / 行 58: @@
 \end{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}
-\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}
-\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}|
-| 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}
-\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}|||
 | 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}
-\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}|
-| 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}}}\\
+|\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}
+|\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}
-\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}
-|\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\\
-|\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}|
 | 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}
-&= \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 ===
 ** The General Formula **
-They are definitions.
+They are definitions?
 ** The formula for N-N Scattering **
@@ 行 105: / 行 112: @@
 ** 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}
-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}
-By using the following equations
+From Quantity 4, the velocity of the c.m. is
 \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}
-$W$ can be written as
+By substituting this formula into the formula of $\boldsymbol{p}_1^*$,
 \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\\
-W^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}
@@ 行 151: / 行 133: @@
 \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}
@@ 行 167: / 行 148: @@
 For the elastic scattering, $m_1 = m_3$ and $m_2 = m_4$. Therefore,
 \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}
@@ 行 178: / 行 156: @@
 \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}
@@ 行 184: / 行 162: @@
 ** 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}
-\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}
 If the second particle is not moving, $\boldsymbol{p}_2=0$ and $E_2=m_2$. Therefore,
 \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}
-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}
-\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}
@@ 行 204: / 行 182: @@
 ** The General Formula **
-In general,
+In the non-relativistic limit,
 \begin{align}
-\gamma = \frac{1}{\sqrt{1-|\boldsymbol{\beta}|^2}}.
+\gamma
+&= \frac{1}{\sqrt{1-|\boldsymbol{\beta}|^2}}\\
+&\approx 1?
 \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 ===
@@ 行 232: / 行 195: @@
 \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}
@@ 行 239: / 行 202: @@
 \begin{align}
 \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}
@@ 行 255: / 行 218: @@
 \begin{align}
 \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}
@@ 行 266: / 行 229: @@
 \begin{align}
 |\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}
 Therefore,
@@ 行 277: / 行 240: @@
 \begin{align}
 \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}
-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 ($\theta_{\rm cm} \rightarrow \theta_{\rm lab}$) ===
@@ 行 287: / 行 250: @@
 \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}
@@ 行 307: / 行 257: @@
 \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}
-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}
-\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}^*.
 \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}
-\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}
@@ 行 325: / 行 274: @@
 \begin{align}
 \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}
@@ 行 334: / 行 283: @@
 \begin{align}
-|\boldsymbol{\beta}_3^*| = |\boldsymbol{\beta}_1^*|\\
+|\boldsymbol{v}_3^*| = |\boldsymbol{v}_1^*|\\
 \delta_{23}^* = \delta_{21}^*
 \end{align}
@@ 行 342: / 行 291: @@
 \begin{align}
 \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}
@@ 行 349: / 行 298: @@
 \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}
@@ 行 369: / 行 305: @@
 \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}
-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}
-\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}
-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}
-\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}
@@ 行 387: / 行 322: @@
 \begin{align}
-\tan\theta_4 = \frac{1}{\gamma_2^*}\cot\frac{\theta_3^*}{2}
+\tan\theta_4 = \cot\frac{\theta_3^*}{2}
 \end{align}
@@ 行 397: / 行 332: @@
 \begin{align}
 \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}
@@ 行 406: / 行 341: @@
 From an equation in the derivation of Quantity 7.,
 \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}
@@ 行 412: / 行 347: @@
 \begin{align}
 \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}
@@ 行 424: / 行 359: @@
 is
 \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}
 Therefore,
 \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}
@@ 行 435: / 行 370: @@
 \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}
@@ 行 455: / 行 377: @@
 \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}
@@ 行 472: / 行 393: @@
 By using the Quantity 7. and $(f/g)'=(f'g-fg')/g^2$,
 \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}
@@ 行 490: / 行 408: @@
 From Quantity 7.,
 \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}
 In general, $\sin\theta=\sqrt{1-\cos^2\theta}$. Therefore,
 \begin{align}
 \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^*}
-&= \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}
 By substituting this formula into the first formula of Quantity 9.,
 \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{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}
@@ 行 515: / 行 436: @@
 \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}
-===  Derivation of Quantity 10. Relations between the $\gamma$ factors ===
+=== Derivation of Quantity 11. Lab quantity relations ===
 ** The first formula of the General Formulae **
-In general,
+From the law of conservation of momentum,
 \begin{align}
-|\boldsymbol{p}|=m\gamma|\boldsymbol{\beta}|=m\sqrt{\gamma^2-1}.
+&&|\boldsymbol{p}_1| = |\boldsymbol{p}_3|\cos\theta_3 + |\boldsymbol{p}_4|\cos\theta_4,\\
+&&|\boldsymbol{p}_3|\sin\theta_3 = |\boldsymbol{p}_4|\sin\theta_4.\\
 \end{align}
-On the other hand, in the Center of mass system,
+By moving $|\boldsymbol{p}_3|\cos\theta_3$ to left and making squares for the both sides of each formula,
 \begin{align}
-|\boldsymbol{p}_1^*|=|\boldsymbol{p}_2^*|.
+\boldsymbol{p}_1^2 + \boldsymbol{p}_3^2\cos^2\theta_3 -2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3 &= \boldsymbol{p}_4^2\cos^2\theta_4,\\
+\boldsymbol{p}_3^2\sin^2\theta_3 &= \boldsymbol{p}_4^2\sin^2\theta_4.\\
 \end{align}
-Therefore,
+By summing these formulae,
 \begin{align}
-m_1\sqrt{\gamma_1^{*2}-1}=m_2\sqrt{\gamma_2^{*2}-1}\\
+\boldsymbol{p}_1^2 + (\boldsymbol{p}_3^2\cos^2\theta_3+\boldsymbol{p}_3^2\sin^2\theta_3) -2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3 &= \boldsymbol{p}_4^2\cos^2\theta_4+\boldsymbol{p}_4^2\sin^2\theta_4,\\
-\Rightarrow m_1^2(\gamma_1^{*2}-1)=m_2^2(\gamma_2^{*2}-1)\\
+\boldsymbol{p}_1^2 + \boldsymbol{p}_3^2 -2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3 &= \boldsymbol{p}_4^2.\\
-\Rightarrow (\gamma_1^{*2}-1)=k_{21}^2(\gamma_2^{*2}-1).
 \end{align}
-** The third formula of the General Formulae **
+(N.B. This formula can be obtained directly by taking square of the both sides of the relation $\boldsymbol{p}_1=\boldsymbol{p}_3+\boldsymbol{p}_4 \Leftrightarrow \boldsymbol{p}_1-\boldsymbol{p}_3=\boldsymbol{p}_4$)
-From Quantity 1. and 5.,
+Then,
 \begin{align}
-\gamma_2^*
+|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3
-&= \gamma_\mathrm{cm}\\
+&= \boldsymbol{p}_1^2 + \boldsymbol{p}_3^2 - \boldsymbol{p}_4^2.
-&= \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,
+From the law of conservation of momentum in the direction of $\boldsymbol{p}_3$,
 \begin{align}
-(\gamma_1^{*2}-1)
+|\boldsymbol{p}_1|\cos\theta_3 &= |\boldsymbol{p}_3|+|\boldsymbol{p}_4|\cos(\theta_3+\theta_4)\\
-&= k_{21}^2\left[\frac{(k_{21}+\gamma_1)^2}{1+k_{21}^2+2\gamma_1k_{21}}-1\right]\\
+\Rightarrow
-&= 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]\\
+|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3
-&= 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}}\\
+&= 2\boldsymbol{p}_3^2+2|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4)\\
-&= k_{21}^2\frac{\gamma_1^2-1}{1+k_{21}^2+2\gamma_1k_{21}}\\
+\Rightarrow
-&= \frac{\gamma_1^2-1}{(1/k_{21})^2+1+2\gamma_1(1/k_{21})}\\
+|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4)
-&= \frac{\gamma_1^2-1}{1+k_{12}^2+2\gamma_1k_{12}}\\
+ &= 2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3-2\boldsymbol{p}_3^2\\
-\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}
+By substituting the first formula of the Quantity 11. Lab quantity relations into this formula,
-Then,
 \begin{align}
-\gamma_1^* = \frac{k_{12}+\gamma_1}{\sqrt{1+k_{12}^2+2\gamma_1k_{12}}}
+|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4)
+&= \boldsymbol{p}_1^2 + \boldsymbol{p}_3^2 - \boldsymbol{p}_4^2 -2\boldsymbol{p}_3^2\\
+&= \boldsymbol{p}_1^2 - \boldsymbol{p}_3^2 - \boldsymbol{p}_4^2
 \end{align}
-** The formulae for N-N Scattering **
+** The formula for N-N Scattering **
-For N-N scattering, $m_1 = m_2 = m_\mathrm{N}$. Therefore,
+From the law of conservation of energy,
 \begin{align}
-k_{12}
+\frac{\boldsymbol{p}_1^2}{2m_1} = \frac{\boldsymbol{p}_3^2}{2m_3} + \frac{\boldsymbol{p}_4^2}{2m_4}
-&= \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,
+For N-N scattering, $m_1 = m_2 = m_3 = m_4 = m_\mathrm{N}$. Therefore,
 \begin{align}
-(\gamma_1^{*2}-1)=k_{21}^2(\gamma_2^{*2}-1)\\
+\boldsymbol{p}_1^2 = \boldsymbol{p}_3^2 + \boldsymbol{p}_4^2.
-\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,
+From this formula and the General formula,
+\begin{align}
+|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4)
+&= \boldsymbol{p}_1^2 - \boldsymbol{p}_3^2 - \boldsymbol{p}_4^2\\
+&= 0
+\end{align}
+Therefore,
 \begin{align}
-\gamma_1^*
+\cos(\theta_3+\theta_4) &= 0\\
-&= \frac{k_{12}+\gamma_1}{\sqrt{1+k_{12}^2+2\gamma_1k_{12}}}\\
+\theta_3+\theta_4 &= 90^\circ.
-&= \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 12. Maximum K.E. transfer to a stationary particle ===
-=== Derivation of Quantity 11. Lab quantity relations ===
+** The General Formula **
-** The first formula of the General Formulae **
-From the law of conservation of momentum,
 \begin{align}
-&&|\boldsymbol{p}_1| = |\boldsymbol{p}_3|\cos\theta_3 + |\boldsymbol{p}_4|\cos\theta_4,\\
+|\boldsymbol{v}_4|\cos\theta_4 &= -|\boldsymbol{v}_4^*|\cos\theta_4^*+v_{\rm cm}\\
-&&|\boldsymbol{p}_3|\sin\theta_3 = |\boldsymbol{p}_4|\sin\theta_4.\\
+|\boldsymbol{v}_4|\sin\theta_4 &= |\boldsymbol{v}_4^*|\sin\theta_4^*
 \end{align}
-By moving $|\boldsymbol{p}_3|\cos\theta_3$ to left and making squares for the both sides of each formula,
+By taking the squares for the both sides,
 \begin{align}
-\boldsymbol{p}_1^2 + \boldsymbol{p}_3^2\cos^2\theta_3 -2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3 &= \boldsymbol{p}_4^2\cos^2\theta_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{p}_3^2\sin^2\theta_3 &= \boldsymbol{p}_4^2\sin^2\theta_4.\\
+|\boldsymbol{v}_4|^2\sin^2\theta_4 &= |\boldsymbol{v}_4^*|^2\sin^2\theta_4^*.
 \end{align}
-By summing these formulae,
+By adding each side,
 \begin{align}
-\boldsymbol{p}_1^2 + (\boldsymbol{p}_3^2\cos^2\theta_3+\boldsymbol{p}_3^2\sin^2\theta_3) -2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3 &= \boldsymbol{p}_4^2\cos^2\theta_4+\boldsymbol{p}_4^2\sin^2\theta_4,\\
+|\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^*\\
-\boldsymbol{p}_1^2 + \boldsymbol{p}_3^2 -2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3 &= \boldsymbol{p}_4^2.\\
+\Rightarrow |\boldsymbol{v}_4|^2 &= |\boldsymbol{v}_4^*|^2+v_{\rm cm}^2-2v_{\rm cm}|\boldsymbol{v}_4^*|\cos\theta_4^*
 \end{align}
-(N.B. This formula can be obtained directly by taking square of the both sides of the relation $\boldsymbol{p}_1=\boldsymbol{p}_3+\boldsymbol{p}_4 \Leftrightarrow \boldsymbol{p}_1-\boldsymbol{p}_3=\boldsymbol{p}_4$)
+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,
-Then,
 \begin{align}
-|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3
+T_1 &= T_3 + T_4\\
-&= \boldsymbol{p}_1^2 + \boldsymbol{p}_3^2 - \boldsymbol{p}_4^2\\
+\boldsymbol{p}_1 &= \boldsymbol{p}_3 + \boldsymbol{p}_4.
-&= 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,
+In general, $T = \frac{\boldsymbol{p}^2}{2m}$. Therefore, the first equation is written as
 \begin{align}
-E_1+m_2&=E_3+E_4,\\
+\frac{\boldsymbol{p}_1^2}{2m_1} &= \frac{\boldsymbol{p}_3^2}{2m_3} + \frac{\boldsymbol{p}_4^2}{2m_4}\\
-E_4&=E_1+m_2-E_3,\\
+\Rightarrow \frac{\boldsymbol{p}_1^2}{m_1} &= \frac{\boldsymbol{p}_3^2}{m_3} + \frac{\boldsymbol{p}_4^2}{m_4}
-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,
+By substituting $\boldsymbol{p}_3 = \boldsymbol{p}_1 - \boldsymbol{p}_4$ into this formula,
 \begin{align}
-|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3
+\frac{\boldsymbol{p}_1^2}{m_1}
-&= m_4^2 - m_1^2 - m_2^2 - m_3^2+2(E_1+m_2)E_3-2E_1m_2.\\
+&= \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}
-** The second formula of the General Formulae **
+Therefore, by using $\boldsymbol{p}^2=|\boldsymbol{p}|^2$,
-From the law of conservation of momentum in the direction of $\boldsymbol{p}_3$,
 \begin{align}
-|\boldsymbol{p}_1|\cos\theta_3 &= |\boldsymbol{p}_3|+|\boldsymbol{p}_4|\cos(\theta_3+\theta_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
-|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3
-&= 2\boldsymbol{p}_3^2+2|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4)\\
-\Rightarrow
-|\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_3-2E_3^2+2m_3^2.\\
 \end{align}
-By substituting the first formula of the Quantity 11. Lab quantity relations into this formula,
+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}
-|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4)
+|\boldsymbol{p}_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\\
+&=\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}}\\
-&= 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\\
+&=\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}
-By substituting the $E_1+m_2=E_3+E_4$ into the formula,
+The direction of $\boldsymbol{p}_4$ is the same as $\boldsymbol{p}_1$, then
 \begin{align}
-|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4)
+\boldsymbol{p}_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\\
+&=\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.
-&= m_3^2 + m_4^2 - m_1^2 - m_2^2-2E_1m_2+2E_3E_4\\
 \end{align}
-** The formula for N-N Scattering **
+From $\boldsymbol{p}_3=\boldsymbol{p}_1-\boldsymbol{p}_4$,
+\begin{align}
+\boldsymbol{p}_3
+&= \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}
-For N-N scattering, $m_1 = m_2 = m_3 = m_4 = m_\mathrm{N}$. Therefore, the second formula of the General Formulae becomes
+By using $T_4=\frac{|\boldsymbol{p}_4|^2}{2m_4}$ and $T_1=\frac{|\boldsymbol{p}_1|^2}{2m_1}$,
 \begin{align}
-|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4)
+T_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\\
+&=\frac{|\boldsymbol{p}_4|^2}{2m_4}\\
-&= m_\mathrm{N}^2 + m_\mathrm{N}^2 - m_\mathrm{N}^2 - m_\mathrm{N}^2-2E_1m_\mathrm{N}+2E_3E_4\\
+&=\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\\
-&= -2E_1m_\mathrm{N}+2E_3E_4\\
+&=\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}\\
-\Rightarrow
+&=\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\\
-|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4)
+&=\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\\
-&= -E_1m_\mathrm{N}+E_3E_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.
-&= E_3E_4-m_\mathrm{N}(E_1+m_\mathrm{N})+m_\mathrm{N}^2\\
 \end{align}
-From the law of conservation of energy, $E_1+m_\mathrm{N}=E_3+E_4$. Therefore,
+As a result,
 \begin{align}
-|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4)
+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.
-&= E_3E_4-m_\mathrm{N}(E_1+m_\mathrm{N})+m_\mathrm{N}^2\\
-&= 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}
+By the way, from $T_3=T_1-T_4$, $T_3$ is
-=== Derivation of Quantity 12. Maximum K.E. transfer to a stationary particle ===
-** The first formula of the General Formulae **
 \begin{align}
-\begin{pmatrix}
+T_3
-E_4\\
+&= T_1-T_4\\
-|\boldsymbol{p}_4|\cos\theta_\mathrm{lab}
+&= 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{pmatrix}
+&= 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\\
-\begin{pmatrix}
+&= 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\\
-\gamma_\mathrm{cm} & \beta_\mathrm{cm}\gamma_\mathrm{cm}\\
+&= \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\\
-\beta_\mathrm{cm}\gamma_\mathrm{cm} & \gamma_\mathrm{cm}
+&= \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\\
-\end{pmatrix}
+&= \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\\
-\begin{pmatrix}
+&= \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\\
-E_4^*\\
+&= \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\\
--|\boldsymbol{p}_4^*|\cos\theta_\mathrm{cm}
+&= \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{pmatrix}\\
-\Rightarrow E_4 &=  \gamma_\mathrm{cm} (E_4^* - \beta_\mathrm{cm}|\boldsymbol{p}_4^*|\cos\theta_\mathrm{cm})
 \end{align}
-Therefore, $E_4$ becomes maximum at $\theta_\mathrm{cm}=\pi$. The maximum value is
+** The first formula for elastic scattering **
+For elastic scattering, $m_3 = m_1$ and $m_4 = m_2$. Therefore,
 \begin{align}
-E_\mathrm{4max} &=  \gamma_\mathrm{cm} (E_4^* + \beta_\mathrm{cm}|\boldsymbol{p}_4^*|).
+\boldsymbol{p}_4&=\frac{2m_2}{m_1+m_2}\boldsymbol{p}_1\\
+\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}
+As a result,
-From this formula,
 \begin{align}
-T_\mathrm{max}
+T_\mathrm{max}=T_4=\frac{4m_1m_2}{(m_1+m_2)^2}T_1
-&= E_\mathrm{4max} - m_4\\
-&= \gamma_\mathrm{cm} (E_4^* + \beta_\mathrm{cm}|\boldsymbol{p}_4^*|) - m_4.
 \end{align}
-For the elastic scattering, $m_4 = m_2$, $|\boldsymbol{p}_4^*| = |\boldsymbol{p}_2^*|$, and $|E_4^*| = |E_2^*|$. Then,
+** 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}
-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\\
-&= \gamma_\mathrm{cm} (E_2^* + \beta_\mathrm{cm}|\boldsymbol{p}_2^*|) - m_2\\
+m_1\boldsymbol{v}_1 = m_1\boldsymbol{v}_3 + m_2\boldsymbol{v}_4\\
-&= \gamma_\mathrm{cm} [m_2\gamma_2^* + \beta_\mathrm{cm}(m_2|\boldsymbol{\beta}_2^*|\gamma_2^*)] - m_2.
 \end{align}
-By using $|\boldsymbol{\beta}_2^*| = \beta_\mathrm{cm}$ and $\gamma_2^* = \gamma_\mathrm{cm}$,
+From the second equation,
 \begin{align}
-T_\mathrm{max}
+\boldsymbol{v}_3 = \frac{m_1\boldsymbol{v}_1 - m_2\boldsymbol{v}_4}{m_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}
-In general, $|\boldsymbol{\beta}| = \sqrt{\gamma^2-1}/\gamma$. Therefore,
-\begin{align}
-T_\mathrm{max}
-&= m_2\gamma_\mathrm{cm}^2 \left(1+\frac{\gamma_\mathrm{cm}^2-1}{\gamma_\mathrm{cm}^2}\right) - m_2\\
-&= 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}
-By using Quantity 4. and 5.,
-\begin{align}
-T_\mathrm{max}
-&= 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}
-$W$ can be written as
+By substituting this formula into the formula of the law of conservation of energy,
 \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}
-** The second formula of the General Formulae **
+** The second formula for elastic scattering **
 From the first formula,
 \begin{align}
 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}
-By using $m_1=m_2/k_{21}$,
+If $m_2/m_1 \ll 1$,
 \begin{align}
 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}
-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}
 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}
-** 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}
 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}
 ==== General memo ====