• TRIUMF Relativistic Handbook, Section V Relativistic Kinematics
  • v_relativistic_kinematics.pdf

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}

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}

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}

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}

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}

The General Formula

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

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, $\theta_\mathrm{3max}=90^\circ$.

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}

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}

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}

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}

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}

\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}

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}