Non-relativistic kinematics [KobaWiki@RCNP]

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===== Non-relativistic kinematics =====
==== References =====
  * TRIUMF Relativistic Handbook, Section V Relativistic Kinematics
    * {{ :research:memos:kinematics:v_relativistic_kinematics.pdf }}

==== Formulae for Non-relativistic Kinematics ====
=== Two Body Kinematics Formulae ===
{{:research:memos:kinematics:lab_cm_systems.gif|}}

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
\begin{align}
\delta_{ij} = |\boldsymbol{v}_i|/|\boldsymbol{v}_j|,
\end{align}
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)**|
| 1. Total c.m. energy (invariant mass) |\begin{align}
W
&= m_1 + m_2\\
&= m_3 + m_4?
\end{align}| Same as the General formula |\begin{align}
W= 2m_\mathrm{N}?
\end{align}|
| 2. c.m. momentum before the interaction |\begin{align}
\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{\boldsymbol{p}_1}{2}
\end{align}|
| 3. c.m. momentum after the interaction |\begin{align}
\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{\boldsymbol{p}_1}{2}
\end{align}| 
| 4. Velocity of the c.m. |\begin{align}
\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}^* \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}{\sqrt{\delta_{23}^{*2}-1}}\\
\mathrm{For\ \ } \delta_{23}^* \ge 1\\
\mathrm{otherwise\ } \theta_{3\mathrm{max}} = 180^\circ
\end{align}| Same as the General formula |\begin{align}
\theta_{3\mathrm{max}} = 90^\circ
\end{align}|
| 7. c.m. to lab angle ($\theta_{\rm cm} \rightarrow \theta_{\rm lab}$) |\begin{align}
\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^*}{\delta_{21}^*+\cos\theta_3^*}\\
\tan\theta_4 &= \cot\frac{\theta_3^*}{2}
\end{align}|\begin{align}
\tan\theta_3 = \frac{\sin\theta_3^*}{1+\cos\theta_3^*}\\
\end{align}|
| 8. lab to c.m. angle transformation ($\theta_\mathrm{lab} \rightarrow \theta_\mathrm{cm}$) |\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}}
\end{align} Another solution (not written in the document) \begin{align}
\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{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^*}(1+\delta_{23}^*\cos\theta_3^*)
\end{align}||\begin{align}
\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. $k_{12}=m_1/m_2$|Undefined?|Undefined?|Undefined?|
| 11. Lab quantity relations |\begin{align}
2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3 &= \boldsymbol{p}_1^2 + \boldsymbol{p}_3^2 - \boldsymbol{p}_4^2\\
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}
|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4) &= 0\\
\theta_3+\theta_4 &= 90^\circ
\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{4m_1m_2}{(m_1+m_2)^2}T_1\\
&= 2m_2 \boldsymbol{v}_\mathrm{cm}^2\\
&\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?

** The formula for N-N Scattering **

Assuming $m_1=m_2=m_\mathrm{N}$,
\begin{align}
W
&= m_1 + m_2\\
&= 2m_\mathrm{N}
\end{align}

=== Derivation of Quantity 2. c.m. momentum before the interaction ===
** The General Formula **

For the center of mass system,
\begin{align}
\boldsymbol{p}_1^* = \boldsymbol{p}_1 - m_1\boldsymbol{v}_\mathrm{cm}.
\end{align}
From Quantity 4, the velocity of the c.m. is
\begin{align}
\boldsymbol{v}_{\rm cm} = \frac{\boldsymbol{p}_1}{m_1+m_2}.
\end{align}
By substituting this formula into the formula of $\boldsymbol{p}_1^*$,
\begin{align}
\boldsymbol{p}_1^*
&= \boldsymbol{p}_1 - m_1\frac{\boldsymbol{p}_1}{m_1+m_2}\\
&= \frac{m_2}{m_1+m_2}\boldsymbol{p}_1\\
&= \frac{m_1m_2}{m_1+m_2}\boldsymbol{v}_1
\end{align}

** The formula for N-N Scattering **

Assuming $m_1=m_2=m_\mathrm{N}$, from the general formula,

\begin{align}
\boldsymbol{p}_1' = 
&= \frac{m_2}{m_1+m_2}\boldsymbol{p}_1\\
&= \frac{m_\mathrm{N}}{2m_\mathrm{N}}\boldsymbol{p}_1\\
&= \frac{\boldsymbol{p}_1}{2}
\end{align}

=== Derivation of Quantity 3. c.m. momentum after the interaction ===
** The General Formula **

It is almost the same as the derivation of Quantity 2.

** The formula for Elastic Scattering **

For the elastic scattering, $m_1 = m_3$ and $m_2 = m_4$. Therefore,
\begin{align}
|\boldsymbol{p}_3^{*}| = |\boldsymbol{p}_1^{*}|
\end{align}

** The formula for N-N Scattering **

For N-N scattering, $m_1 = m_2 = m_3 = m_4 = m_\mathrm{N}$. Therefore,

\begin{align}
|\boldsymbol{p}_3'| = |\boldsymbol{p}_1'| = \frac{\boldsymbol{p}_1}{2}
\end{align}

=== Derivation of Quantity 4. Velocity of the c.m. ===
** The General Formula **

From the notes below, the velocity of the c.m. $\boldsymbol{v}_{\rm cm}$ of two moving particles is written as

\begin{align}
\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{v}_{\rm cm} = \frac{\boldsymbol{p}_1}{m_1+m_2}.
\end{align}

In the center of mass system, $\boldsymbol{v}_\mathrm{cm}=\boldsymbol{v}_2^*$. As a result,

\begin{align}
\boldsymbol{v}_2^* = \boldsymbol{v}_{\rm cm} = \frac{\boldsymbol{p}_1}{m_1+m_2}
\end{align}

=== Derivation of Quantity 5. $\gamma$ of the c.m. ===
** The General Formula **

In the non-relativistic limit,
\begin{align}
\gamma
&= \frac{1}{\sqrt{1-|\boldsymbol{\beta}|^2}}\\
&\approx 1?
\end{align}

=== Derivation of Quantity 6. Maximum lab scattering angle ===
** The General Formula **

From a equation in the derivation of Quantity 7.,

\begin{align}
\tan\theta_3 &= \frac{\sin\theta_3^*}{\delta_{23}^*+\cos\theta_3^*}.\\
\end{align}

By differentiating this equation,

\begin{align}
\frac{d(\tan\theta_3)}{d\theta_3^*}
&= \frac{\cos\theta_3^*[\delta_{23}^*+\cos\theta_3^*]-\sin\theta_3^*(-\sin\theta_3^*)}{\left(\delta_{23}^*+\cos\theta_3^*\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{\delta_{23}^*\cos\theta_3^*+1}{\left(\delta_{23}^*+\cos\theta_3^*\right)^2}.\\
\end{align}

If $\delta_{23}^*< 1$, $\frac{d(\tan\theta_3)}{d\theta_3^*}>0$. Therefore, the $\tan\theta_3$ becomes maximum and reaches zero at $\theta_3^*=180^\circ$. In this case, $\theta_{3\mathrm{max}} = 180^\circ$.

If $\delta_{23}^*\ge 1$ and $\cos\theta_3^*=-1/\delta_{23}^*$, $\frac{d(\tan\theta_3)}{d\theta_3^*}=0$. Therefore, the $\tan\theta_3$ becomes maximum at
\begin{align}
\cos\theta_3^*=-1/\delta_{23}^*.
\end{align}

By substituting this formula into the formula of $\tan\theta_3$,

\begin{align}
\tan\theta_3
&= \frac{\sqrt{1-(1/\delta_{23}^*)^2}}{\delta_{23}^*-1/\delta_{23}^*}\\
&= \frac{\sqrt{\delta_{23}^{*2}-1}}{\delta_{23}^{*2}-1}\\
&= \frac{1}{\sqrt{\delta_{23}^{*2}-1}}.\\
\end{align}

** The formula for N-N Scattering **

For N-N scattering, $m_1 = m_2 = m_3 = m_4 = m_\mathrm{N}$, and

\begin{align}
|\boldsymbol{p}_3'| = |\boldsymbol{p}_1'| = |\boldsymbol{p}_2'|\\
|\boldsymbol{v}_3'| = |\boldsymbol{v}_1'| = |\boldsymbol{v}_2'|\\
\end{align}
Therefore,
\begin{align}
\delta_{23}^* = 1
\end{align}

From a equation in the derivation of Quantity 7.,

\begin{align}
\tan\theta_3
&= \frac{\sin\theta_3^*}{\delta_{23}^*+\cos\theta_3^*}.\\
&= \frac{\sin\theta_3^*}{1+\cos\theta_3^*}.
\end{align}

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 ($\theta_{\rm cm} \rightarrow \theta_{\rm lab}$) ===
** The General Formula **

\begin{align}
|\boldsymbol{v}_3|\cos\theta_{\rm lab} &= |\boldsymbol{v}_3^*|\cos\theta_{\rm cm}+v_{\rm cm}\\
|\boldsymbol{v}_3|\sin\theta_{\rm lab} &= |\boldsymbol{v}_3^*|\sin\theta_{\rm cm}
\end{align}

Therefore,

\begin{align}
\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}}{\cos\theta_{\rm cm}+v_{\rm cm}/|\boldsymbol{v}_3^*|}
\end{align}
In the center of mass system, $v_\mathrm{cm}=|\boldsymbol{v}_2^*|$. Therefore, 
\begin{align}
v_{\rm cm}/|\boldsymbol{v}_3^*|
&= |\boldsymbol{v}_2^*|/|\boldsymbol{v}_3^*|\\
&= \delta_{23}^*.
\end{align}
By using this formula, $\theta_\mathrm{cm}=\theta_3^*$, and $\theta_\mathrm{lab}=\theta_3$,
\begin{align}
\tan\theta_3 &= \frac{\sin\theta_3^*}{\delta_{23}^*+\cos\theta_3^*}.\\
\end{align}

By using the relation $\cos\theta=1/\sqrt{1+\tan^2\theta}$,
\begin{align}
\cos\theta_3
&= \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}}
\end{align}

** The first formula for Elastic Scattering **

For elastic scattering, $m_1 = m_3$ and $|\boldsymbol{p}_3^*| = |\boldsymbol{p}_1^*|$. Therefore,

\begin{align}
|\boldsymbol{v}_3^*| = |\boldsymbol{v}_1^*|\\
\delta_{23}^* = \delta_{21}^*
\end{align}

From a equation in the derivation of Quantity 7.,

\begin{align}
\tan\theta_3
&= \frac{\sin\theta_3^*}{\delta_{23}^*+\cos\theta_3^*}.\\
&= \frac{\sin\theta_3^*}{\delta_{21}^*+\cos\theta_3^*}.\\
\end{align}

** The second formula for Elastic Scattering **

\begin{align}
|\boldsymbol{v}_4|\cos\theta_4 &= -|\boldsymbol{v}_4^*|\cos\theta_4^*+v_{\rm cm}\\
|\boldsymbol{v}_4|\sin\theta_4 &= |\boldsymbol{v}_4^*|\sin\theta_4^*
\end{align}

Therefore,

\begin{align}
\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^*}{v_{\rm cm}/|\boldsymbol{v}_4^*|-\cos\theta_4^*}\\
\end{align}

By using $\theta_4^* = \theta_3^*$ and $v_{\rm cm} = |\boldsymbol{v}_2^*|$,
\begin{align}
\tan\theta_4 = \frac{\sin\theta_3^*}{|\boldsymbol{v}_2^*|/|\boldsymbol{v}_4^*|-\cos\theta_3^*}
\end{align}

For elastic scattering, $m_2=m_4$ and $|\boldsymbol{v}_2^*|=|\boldsymbol{v}_4^*|$. Therefore,
\begin{align}
\tan\theta_4 = \frac{\sin\theta_3^*}{1-\cos\theta_3^*}
\end{align}

In general, $\cot\dfrac{x}{2} = \dfrac{\sin x}{1-\cos x}$. Therefore,

\begin{align}
\tan\theta_4 = \cot\frac{\theta_3^*}{2}
\end{align}

** The formula for N-N Scattering **

From an equation in the derivation of Quantity 6., $\delta_{23}^* = 1$.
From this equation and an equation in the derivation of Quantity 7.,

\begin{align}
\tan\theta_3
&= \frac{\sin\theta_3^*}{\delta_{23}^*+\cos\theta_3^*}.\\
&= \frac{\sin\theta_3^*}{1+\cos\theta_3^*}.\\
\end{align}

=== Derivation of Quantity 8. lab to c.m. angle transformation ($\theta_\mathrm{cm} \rightarrow \theta_\mathrm{lab}$) ===
** The General Formula **

From an equation in the derivation of Quantity 7.,
\begin{align}
\tan\theta_3 &= \frac{\sin\theta_3^*}{\delta_{23}^*+\cos\theta_3^*}.\\
\end{align}

By taking the square of the formula,
\begin{align}
\tan^2\theta_3
&= \frac{\sin^2\theta_3^*}{\left(\delta_{23}^*+\cos\theta_3^*\right)^2}\\
&= \frac{1-\cos^2\theta_3^*}{\left(\delta_{23}^*+\cos\theta_3^*\right)^2},\\
&\Rightarrow \tan^2\theta_3\left(\delta_{23}^*+\cos\theta_3^*\right)^2=1-\cos^2\theta_3^*\\
&\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}

In general, the solution of the equation
\begin{align}
ax^2+2bx+c=0
\end{align}
is
\begin{align}
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}^*\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}


** The another solution **

\begin{align}
|\boldsymbol{v}_3^*|\cos\theta_{\rm cm} &= |\boldsymbol{v}_3|\cos\theta_{\rm lab}-v_{\rm cm}\\
|\boldsymbol{v}_3^*|\sin\theta_{\rm cm} &= |\boldsymbol{v}_3|\sin\theta_{\rm lab}
\end{align}

Therefore,

\begin{align}
\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}}{\cos\theta_{\rm lab}-v_{\rm cm}/|\boldsymbol{v}_3|}\\
\end{align}

=== Derivation of Quantity 9. Solid angle transformation (Jacobian) ===

** The first formula of the General Formulae **

\begin{align}
\frac{d\Omega_3}{d\Omega_3^*}
&= \frac{\sin\theta_3d\theta_3d\phi}{\sin\theta_3^*d\theta_3^*d\phi}\\
&= \frac{\sin\theta_3d\theta_3}{\sin\theta_3^*d\theta_3^*}\\
&= \frac{d(\cos\theta_3)}{d(\cos\theta_3^*)}\\
\end{align}
By using the Quantity 7. and $(f/g)'=(f'g-fg')/g^2$,
\begin{align}
\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{\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{\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[\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^*}}\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[\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[\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[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}

** The second formula of the General Formulae **

From Quantity 7.,
\begin{align}
\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{\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^*+\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)}{\sin^2\theta_3^*+\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^*+\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{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^*)\\
\end{align}

** The formula for N-N Scattering **

From an equation in the derivation of Quantity 6., $\delta_{23}^* = 1$.
Therefore, the first formula of the General Formulae of Quantity 9. can be written as

\begin{align}
\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+\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 11. Lab quantity relations ===

** 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{p}_3|\sin\theta_3 = |\boldsymbol{p}_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,
\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{p}_3^2\sin^2\theta_3 &= \boldsymbol{p}_4^2\sin^2\theta_4.\\
\end{align}
By summing these formulae,
\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{p}_1^2 + \boldsymbol{p}_3^2 -2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3 &= \boldsymbol{p}_4^2.\\
\end{align}

(N.B. This formula can be obtained directly by taking square of the both sides of the relation $\boldsymbol{p}_1=\boldsymbol{p}_3+\boldsymbol{p}_4 \Leftrightarrow \boldsymbol{p}_1-\boldsymbol{p}_3=\boldsymbol{p}_4$)

Then,
\begin{align}
2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3
&= \boldsymbol{p}_1^2 + \boldsymbol{p}_3^2 - \boldsymbol{p}_4^2.
\end{align}

** The second formula of the General Formulae **

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)\\
\Rightarrow
2|\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
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\\
\end{align}
By substituting the first formula of the Quantity 11. Lab quantity relations into this formula,
\begin{align}
2|\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 formula for N-N Scattering **

From the law of conservation of energy,

\begin{align}
\frac{\boldsymbol{p}_1^2}{2m_1} = \frac{\boldsymbol{p}_3^2}{2m_3} + \frac{\boldsymbol{p}_4^2}{2m_4}
\end{align}

For N-N scattering, $m_1 = m_2 = m_3 = m_4 = m_\mathrm{N}$. Therefore, 

\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}
2|\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}
\cos(\theta_3+\theta_4) &= 0\\
\theta_3+\theta_4 &= 90^\circ.
\end{align}

=== Derivation of Quantity 12. Maximum K.E. transfer to a stationary particle ===

** The General Formula **

\begin{align}
|\boldsymbol{v}_4|\cos\theta_4 &= -|\boldsymbol{v}_4^*|\cos\theta_4^*+v_{\rm cm}\\
|\boldsymbol{v}_4|\sin\theta_4 &= |\boldsymbol{v}_4^*|\sin\theta_4^*
\end{align}

By taking the squares for the both sides,

\begin{align}
|\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}

By adding each side,

\begin{align}
|\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^*\\
\Rightarrow |\boldsymbol{v}_4|^2 &= |\boldsymbol{v}_4^*|^2+v_{\rm cm}^2-2v_{\rm cm}|\boldsymbol{v}_4^*|\cos\theta_4^*
\end{align}

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}
T_1 &= T_3 + T_4\\
\boldsymbol{p}_1 &= \boldsymbol{p}_3 + \boldsymbol{p}_4.
\end{align}

In general, $T = \frac{\boldsymbol{p}^2}{2m}$. Therefore, the first equation is written as
\begin{align}
\frac{\boldsymbol{p}_1^2}{2m_1} &= \frac{\boldsymbol{p}_3^2}{2m_3} + \frac{\boldsymbol{p}_4^2}{2m_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}

By substituting $\boldsymbol{p}_3 = \boldsymbol{p}_1 - \boldsymbol{p}_4$ into this formula, 

\begin{align}
\frac{\boldsymbol{p}_1^2}{m_1}
&= \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}

Therefore, by using $\boldsymbol{p}^2=|\boldsymbol{p}|^2$,

\begin{align}
\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\\
\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,
\begin{align}
|\boldsymbol{p}_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}

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}
\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}

By using $T_4=\frac{|\boldsymbol{p}_4|^2}{2m_4}$ and $T_1=\frac{|\boldsymbol{p}_1|^2}{2m_1}$,

\begin{align}
T_4
&=\frac{|\boldsymbol{p}_4|^2}{2m_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\\
&=\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}

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  

\begin{align}
T_3
&= T_1-T_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}

** The first formula for elastic scattering **

For elastic scattering, $m_3 = m_1$ and $m_4 = m_2$. Therefore,
\begin{align}
\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,

\begin{align}
T_\mathrm{max}=T_4=\frac{4m_1m_2}{(m_1+m_2)^2}T_1
\end{align}



** 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}
\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}

From the second equation,
\begin{align}
\boldsymbol{v}_3 = \frac{m_1\boldsymbol{v}_1 - m_2\boldsymbol{v}_4}{m_1}.
\end{align}

By substituting this formula into the formula of the law of conservation of energy,

\begin{align}
\frac{1}{2}m_1\boldsymbol{v}_1^2
&= \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\\
&= \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\\
\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 \\
\Rightarrow 0 &= \frac{m_2|\boldsymbol{v}_4| - 2m_1|\boldsymbol{v}_1|}{m_1} + |\boldsymbol{v}_4| \\
\Rightarrow 2|\boldsymbol{v}_1| &= \left(\frac{m_2}{m_1}+1\right)|\boldsymbol{v}_4|\\
\Rightarrow |\boldsymbol{v}_4|
&= 2\frac{1}{\frac{m_2}{m_1}+1}|\boldsymbol{v}_1|\\
&= \frac{2m_1}{m_1+m_2}|\boldsymbol{v}_1|\\
\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}

** The second formula for elastic scattering **

From the first formula,
\begin{align}
T_\mathrm{max}
&=\frac{4m_1m_2}{(m_1+m_2)^2}T_1\\
&=\frac{2m_2\boldsymbol{p}_1^2}{(m_1+m_2)^2}\\
&=\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}
If $m_2/m_1 \ll 1$,
\begin{align}
T_\mathrm{max}
&=\frac{2m_2\boldsymbol{v}_1^2}{(1+m_2/m_1)^2}\\
&\approx 2m_2\boldsymbol{v}_1^2.\\
\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}

** The formula for N-N scattering **

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}
\boldsymbol{p}^*
&=m(\boldsymbol{v}-\boldsymbol{v}_0)\\
&=\boldsymbol{p}-m\boldsymbol{v}_0
\end{align}

So, this equation can be written in another way.

\begin{align}
\begin{pmatrix}
m\\
\boldsymbol{p}^*
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
-\boldsymbol{v}_0 & 1\\
\end{pmatrix}
\begin{pmatrix}
m\\
\boldsymbol{p}
\end{pmatrix}
\end{align}

=== Velocity of the center of mass $\boldsymbol{v}_\mathrm{cm}$ ===

In the center of mass frame, the sum of the momenta of the particles ($\boldsymbol{p}_1^* + \boldsymbol{p}_2^*$) should be $0$.
Therefore, $\boldsymbol{v}_{\rm cm}$ can be obtained by solving the following equation for $\boldsymbol{v}_0$.
\begin{align}
0 &= \boldsymbol{p}_1^* + \boldsymbol{p}_2^*\\
  &= \boldsymbol{p}_1-m_1\boldsymbol{v}_0 + \boldsymbol{p}_2-m_2\boldsymbol{v}_0\\
\end{align}
By solving the equation for $\boldsymbol{v}_0$,
\begin{align}
\boldsymbol{v}_0 = \frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{m_1+m_2}.
\end{align}
Therefore, the $\boldsymbol{v}_\mathrm{cm}$ is
\begin{align}
\boldsymbol{v}_\mathrm{cm} 
&= \frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{m_1+m_2}\\
&= \frac{m_1\boldsymbol{v}_1 + m_2\boldsymbol{v}_2}{m_1+m_2}
\end{align}