• TRIUMF Relativistic Handbook, Section V Relativistic Kinematics
  • v_relativistic_kinematics.pdf
• TRIUMF Relativistic Handbook Sec. V, VI, VII, etc.
  • triumf_kinematics_handbook.zip

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 $\boldsymbol{\beta}_i$ ($m_i^*$, $\boldsymbol{p}_i^*$, $E_i^*$, and $\boldsymbol{\beta}_i^*$), respectively.

In the following, the quantities $\delta_{ij}$ are defined by \begin{align} \delta_{ij} = |\boldsymbol{\beta}_i|/|\boldsymbol{\beta}_j|, \end{align} where the subscripts refer to the particles.

The General Formula

They are definitions.

The formula for N-N Scattering

Assuming $m_1=m_2=m_\mathrm{N}$, \begin{align} W &= \sqrt{m_1^2+m_2^2+2m_2E_1}\\ &= \sqrt{m_\mathrm{N}^2+m_\mathrm{N}^2+2m_\mathrm{N}E_1}\\ &= \sqrt{2(m_\mathrm{N}^2+m_\mathrm{N}E_1)} \end{align}

The General Formula

In the center of mass frame, $\boldsymbol{p}_1^*+\boldsymbol{p}_2^*=0$. Therefore, \begin{align} W &= \sqrt{(E_1^*+E_2^*)^2-(\boldsymbol{p}_1^*+\boldsymbol{p}_2^*)^2}\\ &= E_1^*+E_2^*. \end{align} By using the following equations \begin{align} |\boldsymbol{p}_1^*| &= |\boldsymbol{p}_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 \begin{align} W &= E_1^*+E_2^*\\ &= \sqrt{\boldsymbol{p}_1^{*2}+m_1^2}+\sqrt{\boldsymbol{p}_1^{*2}+m_2^2} \end{align} By making square for both sides, \begin{align} W^2 &= \boldsymbol{p}_1^{*2}+m_1^2+\boldsymbol{p}_1^{*2}+m_2^2+2\sqrt{\left(\boldsymbol{p}_1^{*2}+m_1^2\right)\left(\boldsymbol{p}_1^{*2}+m_2^2\right)}\\ &= 2\boldsymbol{p}_1^{*2}+m_1^2+m_2^2+2\sqrt{\boldsymbol{p}_1^{*4}+\boldsymbol{p}_1^{*2}\left(m_1^2+m_2^2\right)+m_1^2m_2^2}\\ W^2-2\boldsymbol{p}_1^{*2}-\left(m_1^2+m_2^2\right) &= 2\sqrt{\boldsymbol{p}_1^{*4}+\boldsymbol{p}_1^{*2}\left(m_1^2+m_2^2\right)+m_1^2m_2^2}.\\ \end{align} Again by making square for both sides, \begin{align} \left[W^2-2\boldsymbol{p}_1^{*2}-\left(m_1^2+m_2^2\right)\right]^2 &= 4\left[\boldsymbol{p}_1^{*4}+\boldsymbol{p}_1^{*2}\left(m_1^2+m_2^2\right)+m_1^2m_2^2\right]\\ W^4+4\boldsymbol{p}_1^{*4}+\left(m_1^2+m_2^2\right)^2-4W^2\boldsymbol{p}_1^{*2}-2W^2\left(m_1^2+m_2^2\right)+4\boldsymbol{p}_1^{*2}\left(m_1^2+m_2^2\right)&= 4\boldsymbol{p}_1^{*4}+4\boldsymbol{p}_1^{*2}\left(m_1^2+m_2^2\right)+4m_1^2m_2^2\\ W^4+\left(m_1^2+m_2^2\right)^2-4W^2\boldsymbol{p}_1^{*2}-2W^2\left(m_1^2+m_2^2\right)&= 4m_1^2m_2^2\\ 4W^2\boldsymbol{p}_1^{*2} &= W^4-2W^2\left(m_1^2+m_2^2\right)+\left(m_1^2+m_2^2\right)^2-4m_1^2m_2^2\\ &= W^4-2W^2\left(m_1^2+m_2^2\right)+\left(m_1^2-m_2^2\right)^2\\ &= W^4-W^2\left[\left(m_1+m_2\right)^2+\left(m_1-m_2\right)^2\right]+\left(m_1^2-m_2^2\right)^2\\ &= \left[W^2-\left(m_1+m_2\right)^2\right]\left[W^2-\left(m_1-m_2\right)^2\right]\\ \boldsymbol{p}_1^{*2} &= \frac{1}{4W^2}\left[W^2-\left(m_1+m_2\right)^2\right]\left[W^2-\left(m_1-m_2\right)^2\right]\\ |\boldsymbol{p}_1^{*}| &= \frac{1}{2W}\sqrt{\left[W^2-\left(m_1+m_2\right)^2\right]\left[W^2-\left(m_1-m_2\right)^2\right]}\\ \end{align}

The formula for N-N Scattering

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

\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]}\\ &= \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{1}{2W}\sqrt{\left[W^2-(2m_\mathrm{N})^2\right]W^2}\\ &= \frac{1}{2}\sqrt{W^2-4m_\mathrm{N}^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^{*}| &= \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}

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{1}{2}\sqrt{W^2-4m_\mathrm{N}^2} \end{align}

The General Formula

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

\begin{align} \boldsymbol{\beta}_{\rm cm} = \frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{E_1+E_2}. \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}. \end{align}

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

\begin{align} \boldsymbol{\beta}_2^* = \boldsymbol{\beta}_{\rm cm} = \frac{\boldsymbol{p}_1}{E_1+m_2} \end{align}

The General Formula

In general, \begin{align} \gamma = \frac{1}{\sqrt{1-|\boldsymbol{\beta}|^2}}. \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}

The General Formula

From a 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)}.\\ \end{align}

By differentiating this equation,

\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{\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{\gamma_2^*[\delta_{23}^*\cos\theta_3^*+1]}{\left[\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)\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}}{\gamma_2^*\left(\delta_{23}^*-1/\delta_{23}^*\right)}\\ &= \frac{\sqrt{\delta_{23}^{*2}-1}}{\gamma_2^*\left(\delta_{23}^{*2}-1\right)}\\ &= \frac{1}{\gamma_2^*\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{\beta}_3'| = |\boldsymbol{\beta}_1'| = |\boldsymbol{\beta}_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^*}{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}.\\ &= \frac{\sin\theta_3^*}{\gamma_2^*\left(1+\cos\theta_3^*\right)}. \end{align}

Therefore, $\tan\theta_3$ becomes infinite at $\cos\theta_3^*=-1$. In that case, $\mathrm{3max}=90^\circ$.

The General Formula

\begin{align} \begin{pmatrix} E_3\\ |\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}

Therefore,

\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)}\\ \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}}{\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, \begin{align} \beta_{\rm cm}/|\boldsymbol{\beta}_3^*| &= |\boldsymbol{\beta}_2^*|/|\boldsymbol{\beta}_3^*|\\ &= \delta_{23}^*. \end{align} By using this formula, $\gamma_\mathrm{cm}=\gamma_2^*$, $\theta_\mathrm{cm}=\theta_3^*$, and $\theta_\mathrm{lab}=\theta_3$, \begin{align} \tan\theta_3 &= \frac{\sin\theta_3^*}{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}.\\ \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^*}{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}\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}} \end{align}

Another solution of the general Formula

The formula is written in the derivation of “The General Formula” and “The Second Formula for Elastic Scattering” \begin{align} \tan\theta_3 &= \frac{\sin\theta_3^*}{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)},\\ \tan\theta_4 &= \frac{\sin\theta_4^*}{\gamma_2^*\left(\delta_{24}^*-\cos\theta_4^*\right)}.\\ \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{\beta}_3^*| = |\boldsymbol{\beta}_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^*}{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}.\\ &= \frac{\sin\theta_3^*}{\gamma_2^*\left(\delta_{21}^*+\cos\theta_3^*\right)}.\\ \end{align}

The second formula for Elastic Scattering

\begin{align} \begin{pmatrix} E_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}

Therefore,

\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)}\\ \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^*}{\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 $\gamma_{\rm cm} = \gamma_2^*$, \begin{align} \tan\theta_4 &= \frac{\sin\theta_3^*}{\gamma_2^*\left(|\boldsymbol{\beta}_2^*|/|\boldsymbol{\beta}_4^*|-\cos\theta_3^*\right)}\\ &= \frac{\sin\theta_3^*}{\gamma_2^*\left(\delta_{24}^*-\cos\theta_3^*\right)} \end{align}

For elastic scattering, $m_2=m_4$, $|\boldsymbol{p}_2^*|=|\boldsymbol{p}_4^*|$, $|\boldsymbol{\beta}_2^*|=|\boldsymbol{\beta}_4^*|$, and $\delta_{24}^*=1$. Therefore, \begin{align} \tan\theta_4 = \frac{\sin\theta_3^*}{\gamma_2^*(1-\cos\theta_3^*)} \end{align}

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

\begin{align} \tan\theta_4 = \frac{1}{\gamma_2^*}\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^*}{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}.\\ &= \frac{\sin\theta_3^*}{\gamma_2^*\left(1+\cos\theta_3^*\right)}.\\ \end{align}

The General Formula

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)}.\\ \end{align}

By taking the square of the formula, \begin{align} \tan^2\theta_3 &= \frac{\sin^2\theta_3^*}{\gamma_2^{*2}\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},\\ &\Rightarrow (\gamma_2^*\tan\theta_3)^2\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\\ \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}^*(\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}}. \end{align}

The another solution

\begin{align} \begin{pmatrix} E_3^*\\ |\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}

Therefore,

\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)}\\ \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}}{\gamma_{\rm cm}\left(\cos\theta_{\rm lab}-\beta_{\rm cm}/|\boldsymbol{\beta}_3|\right)}\\ \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{\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]\\ &= \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[\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[\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[\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[\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[\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}\\ &= \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}\\ &= \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}

The second formula of the General Formulae

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}}. \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{\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^*}{\sin^2\theta_3^*+\gamma_2^{*2}\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}}\\ \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}} \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{\sin^3\theta_3}{\sin^3\theta_3^*}\gamma_2^*(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{\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{\gamma_2^*(1+\cos\theta_3^*)}{\left[\sin^2\theta_3^*+\gamma_2^{*2}\left(1+\cos\theta_3^*\right)^2\right]^{3/2}} \end{align}

The first formula of the General Formulae

In general,

\begin{align} |\boldsymbol{p}|=m\gamma|\boldsymbol{\beta}|=m\sqrt{\gamma^2-1}. \end{align} On the other hand, in the Center of mass system, \begin{align} |\boldsymbol{p}_1^*|=|\boldsymbol{p}_2^*|. \end{align} Therefore, \begin{align} m_1\sqrt{\gamma_1^{*2}-1}=m_2\sqrt{\gamma_2^{*2}-1}\\ \Rightarrow m_1^2(\gamma_1^{*2}-1)=m_2^2(\gamma_2^{*2}-1)\\ \Rightarrow (\gamma_1^{*2}-1)=k_{21}^2(\gamma_2^{*2}-1). \end{align}

The third formula of the General Formulae

From Quantity 1. and 5.,

\begin{align} \gamma_2^* &= \gamma_\mathrm{cm}\\ &= \frac{E_1+m_2}{W}\\ &= \frac{E_1+m_2}{\sqrt{m_1^2+m_2^2+2m_2E_1}}\\ &= \frac{m_1\gamma_1+m_2}{\sqrt{m_1^2+m_2^2+2m_1m_2\gamma_1}}. \end{align} Then, by using $k_{12}=m_1/m_2$,

\begin{align} \gamma_2^* = \frac{k_{21}+\gamma_1}{\sqrt{1+k_{21}^2+2\gamma_1k_{21}}} \end{align}

The second formula of the General Formulae

By substituting the third formula into the first formula,

\begin{align} (\gamma_1^{*2}-1) &= k_{21}^2\left[\frac{(k_{21}+\gamma_1)^2}{1+k_{21}^2+2\gamma_1k_{21}}-1\right]\\ &= k_{21}^2\left[\frac{k_{21}^2+\gamma_1^2+2\gamma_1k_{21}}{1+k_{21}^2+2\gamma_1k_{21}}-1\right]\\ &= k_{21}^2\frac{k_{21}^2+\gamma_1^2+2\gamma_1k_{21}-(1+k_{21}^2+2\gamma_1k_{21})}{1+k_{21}^2+2\gamma_1k_{21}}\\ &= k_{21}^2\frac{\gamma_1^2-1}{1+k_{21}^2+2\gamma_1k_{21}}\\ &= \frac{\gamma_1^2-1}{(1/k_{21})^2+1+2\gamma_1(1/k_{21})}\\ &= \frac{\gamma_1^2-1}{1+k_{12}^2+2\gamma_1k_{12}}\\ \Leftrightarrow \gamma_1^{*2} &= \frac{\gamma_1^2-1}{1+k_{12}^2+2\gamma_1k_{12}}+1\\ &= \frac{\gamma_1^2-1+(1+k_{12}^2+2\gamma_1k_{12})}{1+k_{12}^2+2\gamma_1k_{12}}\\ &= \frac{\gamma_1^2+k_{12}^2+2\gamma_1k_{12}}{1+k_{12}^2+2\gamma_1k_{12}}\\ &= \frac{(k_{12}+\gamma_1)^2}{1+k_{12}^2+2\gamma_1k_{12}}\\ \end{align}

Then,

\begin{align} \gamma_1^* = \frac{k_{12}+\gamma_1}{\sqrt{1+k_{12}^2+2\gamma_1k_{12}}} \end{align}

The formulae for N-N Scattering

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

\begin{align} k_{12} &= \frac{m_1}{m_2}\\ &= \frac{m_\mathrm{N}}{m_\mathrm{N}}\\ &= 1\\ \end{align}

Similarly, $k_{21} = 1$. As a result, from the first formula of the General Formulae,

\begin{align} (\gamma_1^{*2}-1)=k_{21}^2(\gamma_2^{*2}-1)\\ \Rightarrow \gamma_1^{*2}-1=\gamma_2^{*2}-1\\ \Rightarrow \gamma_1^*=\gamma_2^*. \end{align}

And, from the second formula of the General Formulae,

\begin{align} \gamma_1^* &= \frac{k_{12}+\gamma_1}{\sqrt{1+k_{12}^2+2\gamma_1k_{12}}}\\ &= \frac{1+\gamma_1}{\sqrt{1+1+2\gamma_1}}\\ &= \frac{1+\gamma_1}{\sqrt{2(1+\gamma_1)}}\\ &= \sqrt{\frac{1+\gamma_1}{2}}\\ \end{align}

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\\ &= E_1^2 - m_1^2 + E_3^2 - m_3^2 - E_4^2 + m_4^2.\\ \end{align} From the law of conservation of energy, \begin{align} E_1+m_2&=E_3+E_4,\\ E_4&=E_1+m_2-E_3,\\ E_4^2&=E_1^2+m_2^2+E_3^2-2(E_1+m_2)E_3+2E_1m_2.\\ \end{align} By substituting this formula into the above formula, \begin{align} 2|\boldsymbol{p}_1||\boldsymbol{p}_3|\cos\theta_3 &= m_4^2 - m_1^2 - m_2^2 - m_3^2+2(E_1+m_2)E_3-2E_1m_2.\\ \end{align}

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\\ &= 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, \begin{align} 2|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4) &= m_4^2 - m_1^2 - m_2^2 - m_3^2+2(E_1+m_2)E_3-2E_1m_2 -2E_3^2+2m_3^2\\ &= 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\\ \end{align} By substituting the $E_1+m_2=E_3+E_4$ into the formula, \begin{align} 2|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4) &= m_3^2 + m_4^2 - m_1^2 - m_2^2+2(E_3+E_4)E_3-2E_1m_2 -2E_3^2\\ &= 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

For N-N scattering, $m_1 = m_2 = m_3 = m_4 = m_\mathrm{N}$. Therefore, the second formula of the General Formulae becomes

\begin{align} 2|\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4) &= m_3^2 + m_4^2 - m_1^2 - m_2^2+2(E_3+E_4)E_3-2E_1m_2 -2E_3^2\\ &= m_\mathrm{N}^2 + m_\mathrm{N}^2 - m_\mathrm{N}^2 - m_\mathrm{N}^2-2E_1m_\mathrm{N}+2E_3E_4\\ &= -2E_1m_\mathrm{N}+2E_3E_4\\ \Rightarrow |\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4) &= -E_1m_\mathrm{N}+E_3E_4\\ &= E_3E_4-m_\mathrm{N}(E_1+m_\mathrm{N})+m_\mathrm{N}^2\\ \end{align}

From the law of conservation of energy, $E_1+m_\mathrm{N}=E_3+E_4$. Therefore, \begin{align} |\boldsymbol{p}_3||\boldsymbol{p}_4|\cos(\theta_3+\theta_4) &= 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}

The formula for N-N Scattering \begin{align} \begin{pmatrix} E_4\\ |\boldsymbol{p}_4|\cos\theta_\mathrm{lab} \end{pmatrix} &= \begin{pmatrix} \gamma_\mathrm{cm} & \beta_\mathrm{cm}\gamma_\mathrm{cm}\\ \beta_\mathrm{cm}\gamma_\mathrm{cm} & \gamma_\mathrm{cm} \end{pmatrix} \begin{pmatrix} E_4^*\\ -|\boldsymbol{p}_4^*|\cos\theta_\mathrm{cm} \end{pmatrix}\\ \Rightarrow E_4 &= \gamma_\mathrm{cm} (E_4^* - \beta_\mathrm{cm}|\boldsymbol{p}_4^*|\cos\theta_\mathrm{cm}) \end{align}

Therefore, $E_4$ becomes maximum at $\theta_\mathrm{cm}=\pi$. The maximum value is \begin{align} E_\mathrm{4max} &= \gamma_\mathrm{cm} (E_4^* + \beta_\mathrm{cm}|\boldsymbol{p}_4^*|). \end{align}

From this formula, \begin{align} T_\mathrm{max} &= 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, \begin{align} T_\mathrm{max} &= \gamma_\mathrm{cm} (E_2^* + \beta_\mathrm{cm}|\boldsymbol{p}_2^*|) - m_2\\ &= \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 $\gamma_2^* = \gamma_\mathrm{cm}$, \begin{align} T_\mathrm{max} &= \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 \begin{align} W &= \sqrt{m_1^2+m_2^2+2m_2E_1}\\ &= \sqrt{m_1^2+m_2^2+2m_2m_1\gamma_1}\\ &= \sqrt{m_1m_2\left(\frac{m_1}{m_2}+\frac{m_2}{m_1}+2\gamma_1\right)}\\ &= \sqrt{m_1m_2\left(k_{12}+k_{21}+2\gamma_1\right)}.\\ \end{align} By substituting this formula into the former formula, \begin{align} T_\mathrm{max} &= 2m_2 \frac{\boldsymbol{p}_1^2}{m_1m_2\left(k_{12}+k_{21}+2\gamma_1\right)}\\ &= \frac{2\boldsymbol{p}_1^2}{m_1(k_{12}+k_{21}+2\gamma_1)}.\\ \end{align}

The second formula of the General Formulae

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{2m_1^2\boldsymbol{\beta}_1^2\gamma_1^2}{m_1(k_{12}+k_{21}+2\gamma_1)}\\ &= \frac{2m_1\boldsymbol{\beta}_1^2\gamma_1^2}{k_{12}+k_{21}+2\gamma_1}.\\ \end{align} By using $m_1=m_2/k_{21}$, \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{\beta}_1^2\gamma_1^2}{1+k_{21}^2+2\gamma_1k_{21}}.\\ \end{align} If $k_{21} = m_2/m_1 \ll 1$, \begin{align} T_\mathrm{max} &= \frac{2m_2\boldsymbol{\beta}_1^2\gamma_1^2}{1+k_{21}^2+2\gamma_1k_{21}}\\ &\approx 2m_2\boldsymbol{\beta}_1^2\gamma_1^2.\\ \end{align}

Another derivation of the second formula of the General Formulae

If $m_2/m_1 \ll 1$, $|\boldsymbol{\beta}_\mathrm{cm}| \approx |\boldsymbol{\beta}_1|$ and $\gamma_\mathrm{cm} \approx \gamma_1$. Therefore, from the formula above, \begin{align} T_\mathrm{max} &= 2m_2 \boldsymbol{\beta}_\mathrm{cm}^2\gamma_\mathrm{cm}^2\\ &\approx 2m_2\boldsymbol{\beta}_1^2\gamma_1^2.\\ \end{align}

The formula for N-N scattering

For N-N scattering,$m_1=m_2=m_\mathrm{N}$ and $k_{12}=k_{21}=1$. 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}}\\ &= \frac{2m_\mathrm{N}\boldsymbol{\beta}_1^2\gamma_1^2}{1+1+2\gamma_1}\\ &= \frac{m_\mathrm{N}\boldsymbol{\beta}_1^2\gamma_1^2}{\gamma_1+1}\\ &= \frac{m_\mathrm{N}(\gamma_1^2-1)}{\gamma_1+1}\\ &= m_\mathrm{N}(\gamma_1-1)\\ &= E_1-m_\mathrm{N}\\ &= T_1 \end{align}

\begin{align} E^*&=\gamma_0(E-\boldsymbol{p}\cdot\boldsymbol{\beta}_0)\\ \boldsymbol{p}^*&=\boldsymbol{p}+\boldsymbol{\beta}_0\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\boldsymbol{\beta}_0\cdot\boldsymbol{p}-E\right] \end{align}

$\boldsymbol{p}$ can be written as \begin{align} \boldsymbol{p} = p_{||}\boldsymbol{e}_{||} +p_{T}\boldsymbol{e}_T, \end{align} where $\boldsymbol{e}_{||}$ ($\boldsymbol{e}_T$) is a unit vector parallel (perpendicular) to $\boldsymbol{\beta}_0$, and $p_{||}$ ($p_T$) is the component of $\boldsymbol{p}$ parallel (perpendicular) to $\boldsymbol{\beta}_0$. Similarly, \begin{align} \boldsymbol{p}^* = p_{||}^*\boldsymbol{e}_{||} +p_{T}^*\boldsymbol{e}_T. \end{align} From the equation of $\boldsymbol{p}$, \begin{align} \boldsymbol{\beta}_0\cdot\boldsymbol{p} &= (\beta_0 \boldsymbol{e}_{||}) \cdot (p_{||}\boldsymbol{e}_{||} +p_{T}\boldsymbol{e}_T)\\ &= \beta_0 p_{||} \boldsymbol{e}_{||} \cdot \boldsymbol{e}_{||} + \beta_0 p_{T} \boldsymbol{e}_{||} \cdot \boldsymbol{e}_T\\ &= \beta_0 p_{||}. \end{align}

By using these equations, \begin{align} E^* &=\gamma_0(E-\boldsymbol{p}\cdot\boldsymbol{\beta}_0)\\ &=\gamma_0(E-p_{||}\beta_0)\\ \boldsymbol{p}^*&=\boldsymbol{p}+\boldsymbol{\beta}_0\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\boldsymbol{\beta}_0\cdot\boldsymbol{p}-E\right]\\ p_{||}^*\boldsymbol{e}_{||} +p_{T}^*\boldsymbol{e}_T &= p_{||}\boldsymbol{e}_{||} +p_{T}\boldsymbol{e}_T + \beta_0 \boldsymbol{e}_{||}\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\beta_0 p_{||}-E\right]\\ &= p_{||}\boldsymbol{e}_{||} +p_{T}\boldsymbol{e}_T + \left[\frac{(\gamma_0\beta_0)^2}{\gamma_0+1} p_{||}-\beta_0 \gamma_0 E\right]\boldsymbol{e}_{||}\\ &= p_{||}\boldsymbol{e}_{||} +p_{T}\boldsymbol{e}_T + \left[\frac{\gamma_0^2-1}{\gamma_0+1} p_{||}-\beta_0 \gamma_0 E\right]\boldsymbol{e}_{||}\\ &= p_{||}\boldsymbol{e}_{||} +p_{T}\boldsymbol{e}_T + \left[(\gamma_0-1) p_{||}-\beta_0 \gamma_0 E\right]\boldsymbol{e}_{||}\\ &= \left(\gamma_0 p_{||}-\beta_0 \gamma_0 E\right)\boldsymbol{e}_{||} +p_{T}\boldsymbol{e}_T.\\ \end{align} Finally, \begin{align} \begin{pmatrix} E^*\\ p_{||}^*\\ p_T^*\\ \end{pmatrix} = \begin{pmatrix} \gamma_0 & -\beta_0\gamma_0 & 0 \\ -\beta_0\gamma_0 & \gamma_0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} E\\ p_{||}\\ p_T\\ \end{pmatrix} \end{align}

The special case where $\boldsymbol{\beta}_0 = \beta_0 \boldsymbol{k}$, with $\boldsymbol{k}$ the unit vector along the $z$-direction, can be written

\begin{align} E^*&=\gamma_0E-\beta_0\gamma_0p_z\\ \boldsymbol{p}^*&=\boldsymbol{p}+\beta_0\boldsymbol{k}\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\beta_0 \boldsymbol{k}\cdot\boldsymbol{p}-E\right]\\ &=\boldsymbol{p}+\beta_0\boldsymbol{k}\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\beta_0 p_z-E\right]\\ &=\boldsymbol{p}+\left[\frac{(\beta_0\gamma_0)^2}{\gamma_0+1}p_z-\beta_0\gamma_0E\right]\boldsymbol{k}\\ &=\boldsymbol{p}+\left[\frac{\gamma_0^2-1}{\gamma_0+1}p_z-\beta_0\gamma_0E\right]\boldsymbol{k}\\ &=\boldsymbol{p}+\left[(\gamma_0-1)p_z-\beta_0\gamma_0E\right]\boldsymbol{k}\\ p_x^*\boldsymbol{i}+p_y^*\boldsymbol{j}+p_z^*\boldsymbol{k}&=p_x\boldsymbol{i}+p_y\boldsymbol{j}+p_z\boldsymbol{k}+\left[(\gamma_0-1)p_z-\beta_0\gamma_0E\right]\boldsymbol{k}\\ p_x^*\boldsymbol{i}+p_y^*\boldsymbol{j}+p_z^*\boldsymbol{k}&=p_x\boldsymbol{i}+p_y\boldsymbol{j}+(\gamma_0 p_z-\beta_0\gamma_0E)\boldsymbol{k}\\ p_x^* &= p_x \\ p_y^* &= p_y \\ p_z^* &= \gamma_0 p_z-\beta_0\gamma_0E \end{align}

\begin{align} \begin{pmatrix} E^*\\ p_x^*\\ p_y^*\\ p_z^* \end{pmatrix} = \begin{pmatrix} \gamma_0 & 0 & 0 & -\beta_0\gamma_0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\beta_0\gamma_0 & 0 & 0 & \gamma_0 \end{pmatrix} \begin{pmatrix} E\\ p_x\\ p_y\\ p_z \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, $\boldsymbol{\beta}_{\rm cm}$ can be obtained by solving the following equation for $\boldsymbol{\beta}_0$. \begin{align} 0 &= \boldsymbol{p}_1^* + \boldsymbol{p}_2^*\\ &= \boldsymbol{p}_1+\boldsymbol{\beta}_0\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\boldsymbol{\beta}_0\cdot\boldsymbol{p}_1-E_1\right] + \boldsymbol{p}_2+\boldsymbol{\beta}_0\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\boldsymbol{\beta}_0\cdot\boldsymbol{p}_2-E_2\right]\\ \end{align} The equation can be rearranged as below. \begin{align} \boldsymbol{p}_1 + \boldsymbol{p}_2 = -\boldsymbol{\beta}_0\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1+\boldsymbol{p}_2)-(E_1+E_2)\right] \end{align} This equation means that the direction of $\boldsymbol{\beta}_0$ is the same as that of $(\boldsymbol{p}_1 + \boldsymbol{p}_2)$. (Note that $\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1+\boldsymbol{p}_2)$ is a scalar term and has no direction.) Therefore, the unit vector $\widehat{\boldsymbol{\beta}_0}$ along the direction of $\boldsymbol{\beta}_0$ is \begin{align} \widehat{\boldsymbol{\beta}_0} = \frac{\boldsymbol{\beta}_0}{\beta_0} = \frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}. \end{align} The magnitude $\beta_0=|\boldsymbol{\beta}_0|$ of the vector $\boldsymbol{\beta}_0$ can be calculated by making inner product of $\boldsymbol{\beta}_0$ and $(\boldsymbol{p}_1 + \boldsymbol{p}_2)$ as below. \begin{align} \boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2) &= -\boldsymbol{\beta}_0\cdot\boldsymbol{\beta}_0\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1+\boldsymbol{p}_2)-(E_1+E_2)\right]\\ &= -\beta_0^2\gamma_0\left[\frac{\gamma_0}{\gamma_0+1}\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1+\boldsymbol{p}_2)-(E_1+E_2)\right]\\ &= -\frac{(\beta_0\gamma_0)^2}{\gamma_0+1}\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1+\boldsymbol{p}_2)+\beta_0^2\gamma_0(E_1+E_2)\\ \beta_0^2\gamma_0(E_1+E_2) &= \left[1+\frac{(\beta_0\gamma_0)^2}{\gamma_0+1}\right]\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2)\\ &= \left[1+\frac{\gamma_0^2-1}{\gamma_0+1}\right]\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2)\\ &= \left[1+(\gamma_0-1)\right]\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2)\\ &= \gamma_0\boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2)\\ \beta_0^2(E_1+E_2) &= \boldsymbol{\beta}_0\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2)\\ \beta_0(E_1+E_2) &= \left(\frac{\boldsymbol{\beta}_0}{\beta_0}\right)\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2)\\ &= \left(\frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\right)\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2)\\ &= \frac{(\boldsymbol{p}_1 + \boldsymbol{p}_2)\cdot(\boldsymbol{p}_1 + \boldsymbol{p}_2)}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\\ &= \frac{|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\\ &=|\boldsymbol{p}_1 + \boldsymbol{p}_2|\\ \beta_0&=\frac{|\boldsymbol{p}_1 + \boldsymbol{p}_2|}{E_1+E_2}. \end{align} Therefore, \begin{align} \widehat{\boldsymbol{\beta}_{\rm cm}} = \frac{\boldsymbol{\beta}_{\rm cm}}{\beta_{\rm cm}} = \frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|},\\ \beta_{\rm cm} = \frac{|\boldsymbol{p}_1 + \boldsymbol{p}_2|}{E_1+E_2}. \end{align} As a result, \begin{align} \boldsymbol{\beta}_{\rm cm} = \beta_{\rm cm}\widehat{\boldsymbol{\beta}_{\rm cm}} = \frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{E_1+E_2}. \end{align} In other words, \begin{align} \boldsymbol{\beta}_{\rm cm} = \frac{E_1\boldsymbol{\beta}_1 + E_2\boldsymbol{\beta}_2}{E_1+E_2}. \end{align}

N.B. In Galilean transformation (non-relativistic limit), \begin{align} \boldsymbol{\beta}_{\rm cm} = \frac{m_1\boldsymbol{\beta}_1 + m_2\boldsymbol{\beta}_2}{m_1+m_2}. \end{align}

\begin{align} W &= \sqrt{(E_1+E_2)^2-(\boldsymbol{p}_1+\boldsymbol{p}_2)^2}\\ \gamma_{\rm cm} &= \frac{1}{\sqrt{1-\beta_{\rm cm}^2}}\\ &= \frac{1}{\sqrt{1-\left(\frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{E_1+E_2}\right)^2}}\\ &= \frac{E_1+E_2}{\sqrt{(E_1+E_2)^2-(\boldsymbol{p}_1 + \boldsymbol{p}_2)^2}}\\ &= \frac{E_1+E_2}{W}\\ \boldsymbol{\beta}_{\rm cm}\gamma_{\rm cm} &= \frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{E_1+E_2} \frac{E_1+E_2}{W} \\ &= \frac{\boldsymbol{p}_1 + \boldsymbol{p}_2}{W} \\ \beta_{\rm cm}\gamma_{\rm cm} &= \frac{|\boldsymbol{p}_1 + \boldsymbol{p}_2|}{W} \end{align}

\begin{align} \boldsymbol{p}_1^* &= \boldsymbol{p}_1+\boldsymbol{\beta}_{\rm cm}\gamma_{\rm cm}\left[\frac{\gamma_{\rm cm}}{\gamma_{\rm cm}+1}\boldsymbol{\beta}_{\rm cm}\cdot\boldsymbol{p}_1-E_1\right]\\ &= \boldsymbol{p}_1+\beta_{\rm cm}\widehat{\boldsymbol{\beta}_{\rm cm}}\gamma_{\rm cm}\left[\frac{\gamma_{\rm cm}}{\gamma_{\rm cm}+1}\beta_{\rm cm}\widehat{\boldsymbol{\beta}_{\rm cm}}\cdot\boldsymbol{p}_1-E_1\right]\\ &= \boldsymbol{p}_1+\widehat{\boldsymbol{\beta}_{\rm cm}}\left[\frac{(\beta_{\rm cm}\gamma_{\rm cm})^2}{\gamma_{\rm cm}+1}\widehat{\boldsymbol{\beta}_{\rm cm}}\cdot\boldsymbol{p}_1-\beta_{\rm cm}\gamma_{\rm cm}E_1\right]\\ &= \boldsymbol{p}_1+\widehat{\boldsymbol{\beta}_{\rm cm}}\left[(\gamma_{\rm cm}-1)\widehat{\boldsymbol{\beta}_{\rm cm}}\cdot\boldsymbol{p}_1-\beta_{\rm cm}\gamma_{\rm cm}E_1\right]\\ &= \boldsymbol{p}_1+\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\left[\left(\frac{E_1+E_2}{W}-1\right)\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\cdot\boldsymbol{p}_1-\frac{|\boldsymbol{p}_1+\boldsymbol{p}_2|}{W}E_1\right]\\ &= \boldsymbol{p}_1+\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\left[\frac{E_1+E_2}{W}\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\cdot\boldsymbol{p}_1-\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\cdot\boldsymbol{p}_1-\frac{|\boldsymbol{p}_1+\boldsymbol{p}_2|}{W}E_1\right]\\ &= \boldsymbol{p}_1+\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\left[\frac{E_1+E_2}{W}\frac{\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}-\frac{\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}-\frac{|\boldsymbol{p}_1+\boldsymbol{p}_2|}{W}E_1\right]\\ &= \boldsymbol{p}_1+\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\left[\frac{(E_1+E_2)(\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2)-E_1|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|}-\frac{\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\right]\\ &= \boldsymbol{p}_1+\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\left[\frac{E_1(\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2)+E_2(\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2)-E_1(\boldsymbol{p}_1^2+\boldsymbol{p}_2^2+2\boldsymbol{p}_1\cdot\boldsymbol{p}_2)}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|}-\frac{\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\right]\\ &= \boldsymbol{p}_1+\frac{\boldsymbol{p}_1+\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\left[\frac{\boldsymbol{p}_1^2E_2-\boldsymbol{p}_2^2E_1+(E_2-E_1)\boldsymbol{p}_1\cdot\boldsymbol{p}_2}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|}-\frac{\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|}\right]\\ &= \frac{\left[\boldsymbol{p}_1^2E_2-\boldsymbol{p}_2^2E_1+(E_2-E_1)\boldsymbol{p}_1\cdot\boldsymbol{p}_2\right](\boldsymbol{p}_1+\boldsymbol{p}_2)}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\boldsymbol{p}_1-\frac{(\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2)(\boldsymbol{p}_1+\boldsymbol{p}_2)}{|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ &= \frac{\left[\boldsymbol{p}_1^2E_2-\boldsymbol{p}_2^2E_1+(E_2-E_1)\boldsymbol{p}_1\cdot\boldsymbol{p}_2\right](\boldsymbol{p}_1+\boldsymbol{p}_2)}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\frac{|\boldsymbol{p}_1+\boldsymbol{p}_2|^2\boldsymbol{p}_1-(\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2)\boldsymbol{p}_1-(\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2)\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ &= \frac{\left[\boldsymbol{p}_1^2E_2-\boldsymbol{p}_2^2E_1+(E_2-E_1)\boldsymbol{p}_1\cdot\boldsymbol{p}_2\right](\boldsymbol{p}_1+\boldsymbol{p}_2)}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\frac{(\boldsymbol{p}_1^2+\boldsymbol{p}_2^2+2\boldsymbol{p}_1\cdot\boldsymbol{p}_2)\boldsymbol{p}_1-(\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2)\boldsymbol{p}_1-(\boldsymbol{p}_1^2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2)\boldsymbol{p}_2}{|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ &= \frac{\left[\boldsymbol{p}_1^2E_2-\boldsymbol{p}_2^2E_1+(E_2-E_1)\boldsymbol{p}_1\cdot\boldsymbol{p}_2\right](\boldsymbol{p}_1+\boldsymbol{p}_2)}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\frac{\boldsymbol{p}_2^2\boldsymbol{p}_1-\boldsymbol{p}_1^2\boldsymbol{p}_2+\boldsymbol{p}_1\cdot\boldsymbol{p}_2(\boldsymbol{p}_1-\boldsymbol{p}_2)}{|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ &= \frac{\left[\boldsymbol{p}_1^2E_2-\boldsymbol{p}_2^2E_1+(E_2-E_1)\cdot\frac{1}{2}\left(|\boldsymbol{p}_1+\boldsymbol{p}_2|^2-\boldsymbol{p}_1^2-\boldsymbol{p}_2^2\right)\right](\boldsymbol{p}_1+\boldsymbol{p}_2)}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\frac{\boldsymbol{p}_2^2\boldsymbol{p}_1-\boldsymbol{p}_1^2\boldsymbol{p}_2+\frac{1}{2}\left(|\boldsymbol{p}_1+\boldsymbol{p}_2|^2-\boldsymbol{p}_1^2-\boldsymbol{p}_2^2\right)(\boldsymbol{p}_1-\boldsymbol{p}_2)}{|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ &= \frac{\left[\boldsymbol{p}_1^2E_2-\boldsymbol{p}_2^2E_1-\frac{1}{2}(E_2-E_1)(\boldsymbol{p}_1^2+\boldsymbol{p}_2^2)+\frac{1}{2}(E_2-E_1)|\boldsymbol{p}_1+\boldsymbol{p}_2|^2\right](\boldsymbol{p}_1+\boldsymbol{p}_2)}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\frac{\boldsymbol{p}_2^2\boldsymbol{p}_1-\boldsymbol{p}_1^2\boldsymbol{p}_2-\frac{1}{2}(\boldsymbol{p}_1^2+\boldsymbol{p}_2^2)(\boldsymbol{p}_1-\boldsymbol{p}_2)+\frac{1}{2}|\boldsymbol{p}_1+\boldsymbol{p}_2|^2(\boldsymbol{p}_1-\boldsymbol{p}_2)}{|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ &= \frac{\left[\frac{1}{2}\boldsymbol{p}_1^2E_2-\frac{1}{2}\boldsymbol{p}_2^2E_1+\frac{1}{2}\boldsymbol{p}_1^2E_1-\frac{1}{2}\boldsymbol{p}_2^2E_2+\frac{1}{2}(E_2-E_1)|\boldsymbol{p}_1+\boldsymbol{p}_2|^2\right](\boldsymbol{p}_1+\boldsymbol{p}_2)}{W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\frac{\frac{1}{2}\boldsymbol{p}_2^2\boldsymbol{p}_1-\frac{1}{2}\boldsymbol{p}_1^2\boldsymbol{p}_2-\frac{1}{2}\boldsymbol{p}_1^2\boldsymbol{p}_1+\frac{1}{2}\boldsymbol{p}_2^2\boldsymbol{p}_2+\frac{1}{2}|\boldsymbol{p}_1+\boldsymbol{p}_2|^2(\boldsymbol{p}_1-\boldsymbol{p}_2)}{|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ &= \frac{\left[(E_1+E_2)(\boldsymbol{p}_1^2-\boldsymbol{p}_2^2)+(E_2-E_1)|\boldsymbol{p}_1+\boldsymbol{p}_2|^2\right](\boldsymbol{p}_1+\boldsymbol{p}_2)}{2W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\frac{(\boldsymbol{p}_2^2-\boldsymbol{p}_1^2)(\boldsymbol{p}_1+\boldsymbol{p}_2)+|\boldsymbol{p}_1+\boldsymbol{p}_2|^2(\boldsymbol{p}_1-\boldsymbol{p}_2)}{2|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ &= \frac{\left[(E_1+E_2)(\boldsymbol{p}_1^2-\boldsymbol{p}_2^2)+(E_2-E_1)|\boldsymbol{p}_1+\boldsymbol{p}_2|^2\right](\boldsymbol{p}_1+\boldsymbol{p}_2)+W(\boldsymbol{p}_2^2-\boldsymbol{p}_1^2)(\boldsymbol{p}_1+\boldsymbol{p}_2)}{2W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}+\frac{1}{2}(\boldsymbol{p}_1-\boldsymbol{p}_2)\\ &= \frac{(E_1+E_2-W)(\boldsymbol{p}_1^2-\boldsymbol{p}_2^2)+(E_2-E_1)|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}{2W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}(\boldsymbol{p}_1+\boldsymbol{p}_2)+\frac{1}{2}(\boldsymbol{p}_1-\boldsymbol{p}_2)\\ &= \frac{(E_1+E_2-W)(\boldsymbol{p}_1^2-\boldsymbol{p}_2^2)(\boldsymbol{p}_1+\boldsymbol{p}_2)+(E_2-E_1)|\boldsymbol{p}_1+\boldsymbol{p}_2|^2(\boldsymbol{p}_1+\boldsymbol{p}_2)+W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2(\boldsymbol{p}_1-\boldsymbol{p}_2)}{2W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ &= \frac{\left[(E_1+E_2-W)(\boldsymbol{p}_1^2-\boldsymbol{p}_2^2)+(E_2-E_1+W)|\boldsymbol{p}_1+\boldsymbol{p}_2|^2\right]\boldsymbol{p}_1+\left[(E_1+E_2-W)(\boldsymbol{p}_1^2-\boldsymbol{p}_2^2)+(E_2-E_1-W)|\boldsymbol{p}_1+\boldsymbol{p}_2|^2\right]\boldsymbol{p}_2}{2W|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ &= \frac{\left[(E_1+E_2-W)(E_1^2-m_1^2-E_2^2+m_2^2)+(E_2-E_1+W)\left[(E_1+E_2)^2-W^2\right]\right]\boldsymbol{p}_1+\left[(E_1+E_2-W)(E_1^2-m_1^2-E_2^2+m_2^2)+(E_2-E_1-W)\left[(E_1+E_2)^2-W^2\right]\right]\boldsymbol{p}_2}{2W\left[(E_1+E_2)^2-W^2\right]}\\ &= \frac{\left[(E_1+E_2-W)(E_1^2-m_1^2-E_2^2+m_2^2)+(E_2-E_1+W)(E_1+E_2+W)(E_1+E_2-W)\right]\boldsymbol{p}_1+\left[(E_1+E_2-W)(E_1^2-m_1^2-E_2^2+m_2^2)+(E_2-E_1-W)(E_1+E_2+W)(E_1+E_2-W)\right]\boldsymbol{p}_2}{2W(E_1+E_2+W)(E_1+E_2-W)}\\ &= \frac{\left[(E_1^2-m_1^2-E_2^2+m_2^2)+(E_2-E_1+W)(E_1+E_2+W)\right]\boldsymbol{p}_1+\left[(E_1^2-m_1^2-E_2^2+m_2^2)+(E_2-E_1-W)(E_1+E_2+W)\right]\boldsymbol{p}_2}{2W(E_1+E_2+W)}\\ &= \frac{\left(m_2^2-m_1^2+2E_2W+W^2\right)\boldsymbol{p}_1+\left(m_2^2-m_1^2-2E_1W-W^2\right)\boldsymbol{p}_2}{2W(E_1+E_2+W)}\\ \end{align}

From the laws of conservation of momentum and energy, \begin{align} \boldsymbol{p}_1 + \boldsymbol{p}_2 &= \boldsymbol{p}_3 + \boldsymbol{p}_4,\\ E_1 + E_2 &= E_3 + E_4.\\ \end{align} By making the square of the both side of the equations \begin{align} \boldsymbol{p}_4 &= \boldsymbol{p}_1 + \boldsymbol{p}_2 - \boldsymbol{p}_3,\\ E_4 &= E_1 + E_2 -E_3, \end{align} one can get \begin{align} |\boldsymbol{p}_4|^2 &= |\boldsymbol{p}_1 + \boldsymbol{p}_2|^2 - 2|\boldsymbol{p}_1 + \boldsymbol{p}_2||\boldsymbol{p}_3|\cos\theta_3 + |\boldsymbol{p}_3|^2,\\ E_4^2 &= (E_1 + E_2)^2-2(E_1+E_2)E_3 + E_3^2.\\ \end{align} Using \begin{align} |\boldsymbol{p}_3|^2 &= E_3^2 -m_3^2,\\ |\boldsymbol{p}_4|^2 &= E_4^2 -m_4^2,\\ \end{align} the first equation becomes \begin{align} E_4^2 -m_4^2 &= |\boldsymbol{p}_1 + \boldsymbol{p}_2|^2 - 2|\boldsymbol{p}_1 + \boldsymbol{p}_2||\boldsymbol{p}_3|\cos\theta_3 + E_3^2 -m_3^2\\ \end{align} By eliminating $E_4$ from the two equations, \begin{align} m_4^2 &= (E_1 + E_2)^2-2(E_1+E_2)E_3 -|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2 + 2|\boldsymbol{p}_1 + \boldsymbol{p}_2||\boldsymbol{p}_3|\cos\theta_3 + m_3^2\\ \end{align} Using the square of the invariant mass $W$ \begin{align} W^2 = (E_1 + E_2)^2 - |\boldsymbol{p}_1 + \boldsymbol{p}_2|^2,\\ \end{align} \begin{align} & 2(E_1+E_2)E_3 - (W^2 + m_3^2-m_4^2) = 2\cos\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2||\boldsymbol{p}_3|\\ \Rightarrow & 2(E_1+E_2)E_3 - (W^2 + m_3^2-m_4^2) = 2\cos\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|\sqrt{E_3^2-m_3^2}\\ \Rightarrow & 4(E_1+E_2)^2E_3^2 - 4(E_1+E_2)\left(W^2+m_3^2-m_4^2\right)E_3 + \left(W^2 + m_3^2-m_4^2\right)^2 = 4\cos^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\left[E_3^2-m_3^2\right]\\ \Rightarrow & 4\left[(E_1+E_2)^2 - \cos^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2 \right]E_3^2 - 4(E_1+E_2)\left(W^2+m_3^2-m_4^2\right)E_3 + \left(W^2+m_3^2-m_4^2\right)^2 + 4\cos^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2m_3^2 = 0\\ \Rightarrow & 4\left[1 - \cos^2\theta_3\left(\frac{|\boldsymbol{p}_1 + \boldsymbol{p}_2|}{E_1+E_2}\right)^2 \right]\left(\frac{E_3}{E_1+E_2}\right)^2 - 4\left[\left(\frac{W}{E_1+E_2}\right)^2 + \frac{m_3^2-m_4^2}{(E_1+E_2)^2}\right]\frac{E_3}{E_1+E_2} + \left[\left(\frac{W}{E_1+E_2}\right)^2 + \frac{m_3^2-m_4^2}{(E_1+E_2)^2}\right]^2 + 4\cos^2\theta_3\left(\frac{|\boldsymbol{p}_1 + \boldsymbol{p}_2|}{E_1+E_2}\right)^2\left(\frac{m_3}{E_1+E_2}\right)^2 = 0\\ \end{align} By using \begin{align} \beta_\mathrm{cm} = \frac{|\boldsymbol{p}_1 + \boldsymbol{p}_2|}{E_1+E_2},\\ \gamma_\mathrm{cm} = \frac{1}{\sqrt{1-\beta_\mathrm{cm}^2}}= \frac{E_1+E_2}{W},\\ 1-\beta_\mathrm{cm}^2 = \left(\frac{W}{E_1+E_2}\right)^2,\\ \end{align} \begin{align} 4\left(1 - \beta_\mathrm{cm}^2\cos^2\theta_3\right)\left(\frac{E_3}{E_1+E_2}\right)^2 - 4\left[1-\beta_\mathrm{cm}^2 + \frac{m_3^2-m_4^2}{(E_1+E_2)^2}\right]\frac{E_3}{E_1+E_2} + \left[1-\beta_\mathrm{cm}^2 + \frac{m_3^2-m_4^2}{(E_1+E_2)^2}\right]^2 + 4\beta_\mathrm{cm}^2\cos^2\theta_3\left(\frac{m_3}{E_1+E_2}\right)^2 = 0\\ \end{align}

For the equation $4ax^2-4bx+c=1$, the solution is \begin{align} x=\frac{b\pm\sqrt{b^2-ac}}{2a}. \end{align}

Therefore, \begin{align} \frac{E_3}{E_1+E_2} &= \frac{1-\beta_\mathrm{cm}^2 + \frac{m_3^2-m_4^2}{(E_1+E_2)^2}\pm\sqrt{\left[1-\beta_\mathrm{cm}^2 + \frac{m_3^2-m_4^2}{(E_1+E_2)^2}\right]^2-\left(1-\beta_\mathrm{cm}^2\cos^2\theta_3\right)\left[\left[1-\beta_\mathrm{cm}^2 + \frac{m_3^2-m_4^2}{(E_1+E_2)^2}\right]^2 + 4\beta_\mathrm{cm}^2\cos^2\theta_3\left(\frac{m_3}{E_1+E_2}\right)^2\right]}}{2\left(1-\beta_\mathrm{cm}^2\cos^2\theta_3\right)}\\ \Rightarrow E_3 &= \frac{1-\beta_\mathrm{cm}^2 + \frac{m_3^2-m_4^2}{(E_1+E_2)^2}\pm\sqrt{\left[1-\beta_\mathrm{cm}^2 + \frac{m_3^2-m_4^2}{(E_1+E_2)^2}\right]^2-\left(1-\beta_\mathrm{cm}^2\cos^2\theta_3\right)\left[\left[1-\beta_\mathrm{cm}^2 + \frac{m_3^2-m_4^2}{(E_1+E_2)^2}\right]^2 + 4\beta_\mathrm{cm}^2\cos^2\theta_3\left(\frac{m_3}{E_1+E_2}\right)^2\right]}}{2\left(1-\beta_\mathrm{cm}^2\cos^2\theta_3\right)}(E_1+E_2)\\ &= \frac{1-\beta_\mathrm{cm}^2 + \frac{m_3^2-m_4^2}{(E_1+E_2)^2}\pm\sqrt{\beta_\mathrm{cm}^2\cos^2\theta_3\left[1-\beta_\mathrm{cm}^2 + \frac{m_3^2-m_4^2}{(E_1+E_2)^2}\right]^2 -4\left(1-\beta_\mathrm{cm}^2\cos^2\theta_3\right)\beta_\mathrm{cm}^2\cos^2\theta_3\left(\frac{m_3}{E_1+E_2}\right)^2}}{2\left(1-\beta_\mathrm{cm}^2\cos^2\theta_3\right)}(E_1+E_2)\\ &= \frac{1-\beta_\mathrm{cm}^2 + \frac{m_3^2-m_4^2}{(E_1+E_2)^2}\pm\beta_\mathrm{cm}\cos\theta_3\sqrt{\left[1-\beta_\mathrm{cm}^2 + \frac{m_3^2-m_4^2}{(E_1+E_2)^2}\right]^2 -4\left(1-\beta_\mathrm{cm}^2\cos^2\theta_3\right)\left(\frac{m_3}{E_1+E_2}\right)^2}}{2\left(1-\beta_\mathrm{cm}^2\cos^2\theta_3\right)}(E_1+E_2)\\ \end{align}

By multiplying $(E1+E2)^2$ to the numerator and the denominator,

\begin{align} E_3 &= \frac{W^2 + m_3^2-m_4^2\pm\beta_\mathrm{cm}\cos\theta_3\sqrt{\left(W^2 +m_3^2-m_4^2\right)^2 -4m_3^2\left(W^2+\sin^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right)}}{2\left(W^2+\sin^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right)}(E_1+E_2)\\ &= \frac{W^2 + m_3^2-m_4^2\pm\beta_\mathrm{cm}\cos\theta_3\sqrt{\left[W^2 -(m_3^2+m_4^2)\right]^2 -4m_3^2m_4^2 -4m_3^2\sin^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2}}{2\left(W^2+\sin^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right)}(E_1+E_2)\\ \end{align}

When $\theta_3=0$ or $\pi$, \begin{align} E_3 &= \frac{1-\beta_\mathrm{cm}^2 + \frac{m_3^2-m_4^2}{(E_1+E_2)^2}\pm\beta_\mathrm{cm}\sqrt{\left[1-\beta_\mathrm{cm}^2 + \frac{m_3^2-m_4^2}{(E_1+E_2)^2}\right]^2 -4\left(1-\beta_\mathrm{cm}^2\right)\left(\frac{m_3}{E_1+E_2}\right)^2}}{2\left(1-\beta_\mathrm{cm}^2\right)}(E_1+E_2) \end{align} By multiplying $(E_1+E_2)^2$ to the numerator and the denominator, \begin{align} E_3 &= \frac{W^2 + m_3^2-m_4^2\pm\beta_\mathrm{cm}\sqrt{\left[W^2 -(m_3^2+m_4^2)\right]^2 -4m_3^2m_4^2}}{2W^2}(E_1+E_2)\\ &=\frac{1}{2}\left[ 1 + \frac{m_3^2-m_4^2}{W^2}\pm\beta_\mathrm{cm}\sqrt{\left(1 -\frac{m_3^2+m_4^2}{W^2}\right)^2 -4\frac{m_3^2m_4^2}{W^4}}\right](E_1+E_2)\\ &=\frac{1}{2}\left[ 1 + \frac{m_3^2-m_4^2}{W^2}\pm\beta_\mathrm{cm}\sqrt{1 -2\frac{m_3^2+m_4^2}{W^2} +\frac{(m_3^2-m_4^2)^2}{W^4}}\right](E_1+E_2)\\ \end{align}

At first, solve the equation: \begin{align} ax-b = c \sqrt{x^2-d^2}. \end{align} By making the squares of the both sides, \begin{align} & a^2x^2-2abx+b^2 = c^2(x^2-d^2)\\ & \Rightarrow (a^2-c^2)x^2-2abx +b^2+c^2d^2=0. \end{align} Then, \begin{align} x &= \frac{ab\pm\sqrt{a^2b^2-(a^2-c^2)(b^2+c^2d^2)}}{a^2-c^2}\\ &= \frac{ab\pm\sqrt{a^2b^2-a^2b^2-a^2c^2d^2+b^2c^2+c^4d^2}}{a^2-c^2}\\ &= \frac{ab\pm\sqrt{-a^2c^2d^2+b^2c^2+c^4d^2}}{a^2-c^2}\\ &= \frac{ab{\pm}c\sqrt{-a^2d^2+b^2+c^2d^2}}{a^2-c^2}\\ &= \frac{ab{\pm}c\sqrt{b^2-d^2(a^2-c^2)}}{a^2-c^2}.\\ \end{align} Therefore, from \begin{align} 2(E_1+E_2)E_3 - (W^2 + m_3^2-m_4^2) = 2\cos\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|\sqrt{E_3^2-m_3^2}, \end{align} $E_3$ is \begin{align} E_3 &= \frac{2(E_1+E_2)(W^2 + m_3^2-m_4^2){\pm}2\cos\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|\sqrt{(W^2 + m_3^2-m_4^2)^2-m_3^2\left[4(E_1+E_2)^2-4\cos^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right]}}{4(E_1+E_2)^2-4\cos^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2}\\ &= \frac{(E_1+E_2)(W^2 + m_3^2-m_4^2){\pm}\cos\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|\sqrt{(W^2 + m_3^2-m_4^2)^2-4m_3^2\left[(E_1+E_2)^2-\cos^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right]}}{2(E_1+E_2)^2-2\cos^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2}\\ &= \frac{W^2 + m_3^2-m_4^2{\pm}\beta_\mathrm{cm}\cos\theta_3\sqrt{(W^2 + m_3^2-m_4^2)^2-4m_3^2\left[(E_1+E_2)^2-\cos^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right]}}{2\left[(E_1+E_2)^2-\cos^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right]}(E_1+E_2)\\ &= \frac{W^2 + m_3^2-m_4^2\pm\beta_\mathrm{cm}\cos\theta_3\sqrt{\left(W^2 +m_3^2-m_4^2\right)^2 -4m_3^2\left(W^2+\sin^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right)}}{2\left(W^2+\sin^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right)}(E_1+E_2)\\ \end{align} If $|\boldsymbol{p}_2|=0$ ($E_2=m_2$), \begin{align} E_3 &= \frac{(E_1+m_2)(W^2 + m_3^2-m_4^2){\pm}\cos\theta_3|\boldsymbol{p}_1|\sqrt{(W^2 + m_3^2-m_4^2)^2-4m_3^2\left[(E_1+m_2)^2-\cos^2\theta_3|\boldsymbol{p}_1|^2\right]}}{2(E_1+m_2)^2-2\cos^2\theta_3|\boldsymbol{p}_1|^2}\\ \end{align} If $|\boldsymbol{p}_2|=0$ ($E_2=m_2$) and elastic scattering ($m_3=m_1$, $m_4=m_2$), \begin{align} E_3 &= \frac{(E_1+m_2)(W^2 + m_1^2-m_2^2){\pm}\cos\theta_3|\boldsymbol{p}_1|\sqrt{(W^2 + m_1^2-m_2^2)^2-4m_1^2\left[(E_1+m_2)^2-\cos^2\theta_3|\boldsymbol{p}_1|^2\right]}}{2(E_1+m_2)^2-2\cos^2\theta_3|\boldsymbol{p}_1|^2}\\ &= \frac{(E_1+m_2)(m_1^2+m_2^2+2E_1m_2 + m_1^2-m_2^2){\pm}\cos\theta_3|\boldsymbol{p}_1|\sqrt{(m_1^2+m_2^2+2E_1m_2 + m_1^2-m_2^2)^2-4m_1^2\left[(E_1+m_2)^2-\cos^2\theta_3|\boldsymbol{p}_1|^2\right]}}{2(E_1+m_2)^2-2\cos^2\theta_3|\boldsymbol{p}_1|^2}\\ &= \frac{(E_1+m_2)(2m_1^2+2E_1m_2){\pm}\cos\theta_3|\boldsymbol{p}_1|\sqrt{(2m_1^2+2E_1m_2)^2-4m_1^2\left[(E_1+m_2)^2-\cos^2\theta_3|\boldsymbol{p}_1|^2\right]}}{2(E_1+m_2)^2-2\cos^2\theta_3|\boldsymbol{p}_1|^2}\\ &= \frac{(E_1+m_2)(m_1^2+E_1m_2){\pm}\cos\theta_3|\boldsymbol{p}_1|\sqrt{(m_1^2+E_1m_2)^2-m_1^2\left[(E_1+m_2)^2-\cos^2\theta_3|\boldsymbol{p}_1|^2\right]}}{(E_1+m_2)^2-\cos^2\theta_3|\boldsymbol{p}_1|^2}\\ &= \frac{(E_1+m_2)(m_1^2+E_1m_2){\pm}\cos\theta_3|\boldsymbol{p}_1|\sqrt{m_1^4+2m_1^2m_2E_1+E_1^2m_2^2-m_1^2E_1^2-2m_1^2E_1m_2-m_1^2m_2^2+m_1^2\cos^2\theta_3|\boldsymbol{p}_1|^2}}{(E_1+m_2)^2-\cos^2\theta_3|\boldsymbol{p}_1|^2}\\ &= \frac{(E_1+m_2)(m_1^2+E_1m_2){\pm}\cos\theta_3|\boldsymbol{p}_1|\sqrt{m_1^4+E_1^2m_2^2-m_1^2E_1^2-m_1^2m_2^2+m_1^2\cos^2\theta_3|\boldsymbol{p}_1|^2}}{(E_1+m_2)^2-\cos^2\theta_3|\boldsymbol{p}_1|^2}\\ &= \frac{(E_1+m_2)(m_1^2+E_1m_2){\pm}\cos\theta_3|\boldsymbol{p}_1|\sqrt{(m_1^2-E_1^2)(m_1^2-m_2^2)+m_1^2\cos^2\theta_3|\boldsymbol{p}_1|^2}}{(E_1+m_2)^2-\cos^2\theta_3|\boldsymbol{p}_1|^2}\\ &= \frac{(E_1+m_2)(m_1^2+E_1m_2){\pm}\cos\theta_3|\boldsymbol{p}_1|\sqrt{-|\boldsymbol{p}_1|^2(m_1^2-m_2^2)+m_1^2\cos^2\theta_3|\boldsymbol{p}_1|^2}}{(E_1+m_2)^2-\cos^2\theta_3|\boldsymbol{p}_1|^2}\\ &= \frac{(E_1+m_2)(m_1^2+E_1m_2){\pm}\cos\theta_3|\boldsymbol{p}_1|^2\sqrt{m_2^2+m_1^2(\cos^2\theta_3-1)}}{(E_1+m_2)^2-\cos^2\theta_3|\boldsymbol{p}_1|^2}\\ &= \frac{(E_1+m_2)(m_1^2+E_1m_2){\pm}\cos\theta_3|\boldsymbol{p}_1|^2\sqrt{m_2^2-m_1^2\sin^2\theta_3}}{(E_1+m_2)^2-\cos^2\theta_3|\boldsymbol{p}_1|^2}\\ \end{align}

In general, as shown in the section above, $E_3$ is \begin{align} E_3 = \frac{(E_1+E_2)(W^2 + m_3^2-m_4^2){\pm}\cos\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|\sqrt{(W^2 + m_3^2-m_4^2)^2-4m_3^2\left[(E_1+E_2)^2-\cos^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right]}}{2(E_1+E_2)^2-2\cos^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2}.\\ \end{align} From a similar derivation, $E_4$ is \begin{align} E_4 &= \frac{(E_1+E_2)(W^2 + m_4^2-m_3^2){\pm}\cos\theta_4|\boldsymbol{p}_1 + \boldsymbol{p}_2|\sqrt{(W^2 + m_4^2-m_3^2)^2-4m_3^2\left[(E_1+E_2)^2-\cos^2\theta_4|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right]}}{2(E_1+E_2)^2-2\cos^2\theta_4|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2}\\ \end{align}

If $|\boldsymbol{p}_2| = 0$, $E_2 = m_2$ and elastic scattering ($m_3=m_1$, $m_4=m_2$), \begin{align} E_4 &= \frac{(E_1+m_2)(W^2 + m_2^2-m_1^2){\pm}\cos\theta_4|\boldsymbol{p}_1|\sqrt{(W^2 + m_2^2-m_1^2)^2-4m_2^2\left[(E_1+m_2)^2-\cos^2\theta_4|\boldsymbol{p}_1+\boldsymbol{p}_2|^2\right]}}{2(E_1+m_2)^2-2\cos^2\theta_4|\boldsymbol{p}_1+\boldsymbol{p}_2|^2}\\ &= \frac{(E_1+m_2)(m_1^2+m_2^2+2E_1m_2 + m_2^2-m_1^2){\pm}\cos\theta|\boldsymbol{p}_1|\sqrt{(m_1^2+m_2^2+2E_1m_2 + m_2^2-m_1^2)^2-4m_2^2\left[(E_1+m_2)^2-\cos^2\theta|\boldsymbol{p}_1|^2\right]}}{2(E_1+m_2)^2-2\cos^2\theta_4|\boldsymbol{p}_1|^2}\\ &= \frac{(E_1+m_2)(2m_2^2+2E_1m_2){\pm}\cos\theta_4|\boldsymbol{p}_1|\sqrt{(2m_2^2+2E_1m_2)^2-4m_2^2\left[(E_1+m_2)^2-\cos^2\theta_4|\boldsymbol{p}_1|^2\right]}}{2(E_1+m_2)^2-2\cos^2\theta_4|\boldsymbol{p}_1|^2}\\ &= \frac{(E_1+m_2)(m_2^2+E_1m_2){\pm}\cos\theta_4|\boldsymbol{p}_1|\sqrt{(m_2^2+E_1m_2)^2-m_2^2\left[(E_1+m_2)^2-\cos^2\theta_4|\boldsymbol{p}_1|^2\right]}}{(E_1+m_2)^2-\cos^2\theta_4|\boldsymbol{p}_1|^2}\\ &= \frac{(E_1+m_2)^2{\pm}\cos\theta_4|\boldsymbol{p}_1|\sqrt{(m_2+E_1)^2-\left[(E_1+m_2)^2-\cos^2\theta_4|\boldsymbol{p}_1|^2\right]}}{(E_1+m_2)^2-\cos^2\theta_4|\boldsymbol{p}_1|^2}m_2\\ &= \frac{(E_1+m_2)^2{\pm}\cos^2\theta_4|\boldsymbol{p}_1|^2}{(E_1+m_2)^2-\cos^2\theta_4|\boldsymbol{p}_1|^2}m_2\\ \end{align} $E_4 \ne m_2$, and then \begin{align} E_4 &= \frac{(E_1+m_2)^2+\cos^2\theta_4|\boldsymbol{p}_1|^2}{(E_1+m_2)^2-\cos^2\theta_4|\boldsymbol{p}_1|^2}m_2\\ \end{align}

From the laws of conservation of momentum and energy, \begin{align} \boldsymbol{p}_1 + \boldsymbol{p}_2 &= \boldsymbol{p}_3 + \boldsymbol{p}_4,\\ E_1 + E_2 &= E_3 + E_4.\\ \end{align} Similar with the derivation of $E_3$, \begin{align} & 2(E_1+E_2)E_3 - (W^2 + m_3^2-m_4^2) = 2\cos\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2||\boldsymbol{p}_3|\\ \Rightarrow & 2(E_1+E_2)\sqrt{|\boldsymbol{p}_3|^2+m_3^2} = 2\cos\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2||\boldsymbol{p}_3|+(W^2 + m_3^2-m_4^2).\\ \end{align} For simplicity, solve the following equation. \begin{align} a \sqrt{x^2+b^2} = cx+d. \end{align} By making the squares of the both sides, \begin{align} & a^2(x^2+b^2) = c^2x^2+2cdx+d^2\\ & \Rightarrow (a^2-c^2)x^2-2cdx +a^2b^2-d^2=0. \end{align} Then, \begin{align} x &= \frac{cd\pm\sqrt{c^2d^2-(a^2-c^2)(a^2b^2-d^2)}}{a^2-c^2}\\ &= \frac{cd\pm\sqrt{c^2d^2-a^4b^2+a^2d^2+a^2b^2c^2-c^2d^2}}{a^2-c^2}\\ &= \frac{cd\pm\sqrt{-a^4b^2+a^2d^2+a^2b^2c^2}}{a^2-c^2}\\ &= \frac{cd{\pm}a\sqrt{-a^2b^2+d^2+b^2c^2}}{a^2-c^2}\\ &= \frac{cd{\pm}a\sqrt{d^2+b^2c^2-a^2b^2}}{a^2-c^2}.\\ &= \frac{cd{\pm}a\sqrt{d^2-b^2(a^2-c^2)}}{a^2-c^2}.\\ \end{align} Therefore, \begin{align} |\boldsymbol{p}_3| &= \frac{2\cos\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|(W^2 + m_3^2-m_4^2){\pm}2(E_1+E_2)\sqrt{(W^2 + m_3^2-m_4^2)^2-m_3^2\left[4(E_1+E_2)^2-4\cos^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right]}}{4(E_1+E_2)^2-4\cos^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2}\\ &= \frac{\cos\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|(W^2 + m_3^2-m_4^2){\pm}(E_1+E_2)\sqrt{(W^2 + m_3^2-m_4^2)^2-m_3^2\left[4(E_1+E_2)^2-4\cos^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right]}}{2\left[(E_1+E_2)^2-\cos^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right]}\\ \end{align}

In the center of mass frame, \begin{align} \boldsymbol{p}_3^* + \boldsymbol{p}_4^* = 0,\\ E_3^* + E_4^* = W.\\ \end{align} Therefore, \begin{align} |\boldsymbol{p}_3^*| = |\boldsymbol{p}_4^*|.\\ \end{align} Then, \begin{align} & W- E_3^* = E_4^* \\ & \Rightarrow E_3^{*2} - 2WE_3^* + W^2 = E_4^{*2}\\ & \Rightarrow |\boldsymbol{p}_3^*|^2+m_3^2 -2WE_3^* + W^2 = |\boldsymbol{p}_4^*|^2+m_4^2\\ & \Rightarrow |\boldsymbol{p}_3^*|^2+m_3^2 -2WE_3^* + W^2 = |\boldsymbol{p}_3^*|^2+m_4^2\\ & \Rightarrow m_3^2 -2WE_3^* + W^2 = m_4^2\\ & \Rightarrow 2WE_3^* = W^2 + m_3^2 - m_4^2\\ & \Rightarrow E_3^* = \frac{W^2 + m_3^2 - m_4^2}{2W} \end{align}

$E_4^*$ is \begin{align} E_4^* &= W - E_3^*\\ &= W-\frac{W^2 + m_3^2 - m_4^2}{2W}\\ &= \frac{2W^2 - W^2 - m_3^2 + m_4^2}{2W}\\ &= \frac{W^2 - m_3^2 + m_4^2}{2W}\\ \end{align} Therefore, \begin{align} E_4^* &= \frac{W^2 + m_4^2 - m_3^2}{2W}\\ \end{align}

$|\boldsymbol{p}_3^*|$ and $|\boldsymbol{p}_4^*|$ are \begin{align} |\boldsymbol{p}_3^*| &= \sqrt{E_3^{*2}-m_3^2}\\ &= \sqrt{\left(\frac{W^2 + m_3^2 - m_4^2}{2W}\right)^2-m_3^2}\\ &= \sqrt{\frac{W^4 + m_3^4 + m_4^4 + 2W^2m_3^2 - 2W^2m_4^2 - 2m_3^2m_4^2 - 4W^2m_3^2 }{4W^2}}\\ &= \sqrt{\frac{W^4 + m_3^4 + m_4^4 - 2W^2m_3^2 - 2W^2m_4^2 - 2m_3^2m_4^2}{4W^2}}\\ &= \frac{1}{2W}\sqrt{W^4 + m_3^4 + m_4^4 - 2W^2m_3^2 - 2W^2m_4^2 - 2m_3^2m_4^2}\\ &= \frac{1}{2W}\sqrt{W^4 - 2(m_3^2 + m_4^2)W^2 + (m_3^2-m_4^2)^2}\\ &= \frac{1}{2W}\sqrt{W^4 - \left[(m_3 + m_4)^2+(m_3 - m_4)^2\right]W^2 + (m_3+m_4)^2(m_3-m_4)^2}\\ &= \frac{1}{2W}\sqrt{\left[W^2 - (m_3 + m_4)^2\right]\left[W^2 - (m_3 - m_4)^2\right]}.\\ \end{align}

Therefore,

\begin{align} |\boldsymbol{p}_3^*| = |\boldsymbol{p}_4^*| &= \frac{1}{2W}\sqrt{\left[W^2 - (m_3 + m_4)^2\right]\left[W^2 - (m_3 - m_4)^2\right]}.\\ \end{align}