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research:memos:kinematics:non-relativistic_kinematics [2017/10/02 03:08] – [Formulae for Non-relativistic Kinematics] kobayashresearch:memos:kinematics:non-relativistic_kinematics [2020/07/29 17:10] (現在) – [Formulae for Non-relativistic Kinematics] kobayash
行 8: 行 8:
 {{:research:memos:kinematics:lab_cm_systems.gif|}} {{:research:memos:kinematics:lab_cm_systems.gif|}}
  
-Adopt units where $c=1$. In the Laboratory System (Center of Mass System), mass, momentum, total energy, and velocity of the $i$-th particle are $m_i$, $\boldsymbol{p}_i$, and $E_i$, and $\boldsymbol{v}_i$ ($m_i^*$, $\boldsymbol{p}_i^*$, $E_i^*$, and $\boldsymbol{v}_i^*$), respectively.+Adopt units where $c=1$. In the Laboratory System (Center of Mass System), mass, momentum, kinetic energy, and velocity of the $i$-th particle are $m_i$, $\boldsymbol{p}_i$, and $T_i$, and $\boldsymbol{v}_i$ ($m_i^*$, $\boldsymbol{p}_i^*$, $T_i^*$, and $\boldsymbol{v}_i^*$), respectively.
  
 In the following, the quantities $\delta_{ij}$ are defined by In the following, the quantities $\delta_{ij}$ are defined by
行 14: 行 14:
 \delta_{ij} = |\boldsymbol{v}_i|/|\boldsymbol{v}_j|, \delta_{ij} = |\boldsymbol{v}_i|/|\boldsymbol{v}_j|,
 \end{align} \end{align}
-where the subscripts refer to the particles.+where the subscripts refer to the particles. In the case of elastic scattering, $|\boldsymbol{p}_3^{*}| = |\boldsymbol{p}_1^{*}|$ as below. In addition, $\boldsymbol{p}_1^* = m_1\boldsymbol{v}_1^* = \frac{m_2}{m_1+m_2}\boldsymbol{p}_1$ and $\boldsymbol{v}_2^* = \boldsymbol{v}_{\rm cm} = \frac{\boldsymbol{p}_1} {m_1+m_2}$. From these formula, 
 + 
 +\begin{align} 
 +\delta_{23}^* = \delta_{21}^* = |\boldsymbol{v}_2^*|/|\boldsymbol{v}_1^*| = m_1/m_2. 
 +\end{align}
  
 |** Quantity **|** General Formula **|**Elastic Scattering**|**N-N Scattering (equal mass)**| |** Quantity **|** General Formula **|**Elastic Scattering**|**N-N Scattering (equal mass)**|
行 61: 行 65:
 \tan\theta_3 = \frac{\sin\theta_3^*}{1+\cos\theta_3^*}\\ \tan\theta_3 = \frac{\sin\theta_3^*}{1+\cos\theta_3^*}\\
 \end{align}| \end{align}|
-| 8. lab to c.m. angle transformation ($\theta_\mathrm{cm} \rightarrow \theta_\mathrm{lab}$) |\begin{align} +| 8. lab to c.m. angle transformation ($\theta_\mathrm{lab} \rightarrow \theta_\mathrm{cm}$) |\begin{align} 
-\cos\theta_3^*=\frac{\delta_{23}^*\tan^2\theta_3}{\tan^2\theta_3+1}\pm\sqrt{\left(\frac{\delta_{23}^*\tan^2\theta_3}{\tan^2\theta_3+1}\right)^2-\frac{\delta_{23}^{*2}\tan^2\theta_3-1}{\tan^2\theta_3+1}}+\cos\theta_3^*=-\frac{\delta_{23}^*\tan^2\theta_3}{\tan^2\theta_3+1}\pm\sqrt{\left(\frac{\delta_{23}^*\tan^2\theta_3}{\tan^2\theta_3+1}\right)^2-\frac{\delta_{23}^{*2}\tan^2\theta_3-1}{\tan^2\theta_3+1}}
 \end{align} Another solution (not written in the document) \begin{align} \end{align} Another solution (not written in the document) \begin{align}
 \tan\theta_{\rm cm} = \frac{\sin\theta_{\rm lab}}{\left(\cos\theta_{\rm lab}-v_{\rm cm}/|\boldsymbol{v}_3|\right)} \tan\theta_{\rm cm} = \frac{\sin\theta_{\rm lab}}{\left(\cos\theta_{\rm lab}-v_{\rm cm}/|\boldsymbol{v}_3|\right)}
行 87: 行 91:
 &= 2m_2 \boldsymbol{v}_\mathrm{cm}^2\\ &= 2m_2 \boldsymbol{v}_\mathrm{cm}^2\\
 &\approx 2m_2\boldsymbol{v}_1^2 &\approx 2m_2\boldsymbol{v}_1^2
-\end{align}| Empty |+\end{align}|\begin{align} 
 +T_\mathrm{max}=T_1 
 +\end{align}|
  
 === Derivation of Quantity 1. Total c.m. energy === === Derivation of Quantity 1. Total c.m. energy ===
行 353: 行 359:
 is is
 \begin{align} \begin{align}
-x=\frac{b}{a}\pm\sqrt{\left(\frac{b}{a}\right)^2-\frac{c}{a}}.+x=-\frac{b}{a}\pm\sqrt{\left(\frac{b}{a}\right)^2-\frac{c}{a}}.
 \end{align} \end{align}
 Therefore, Therefore,
 \begin{align} \begin{align}
-\cos\theta_3^*=\frac{\delta_{23}^*\tan^2\theta_3}{\tan^2\theta_3+1}\pm\sqrt{\left(\frac{\delta_{23}^*\tan^2\theta_3}{\tan^2\theta_3+1}\right)^2-\frac{\delta_{23}^{*2}\tan^2\theta_3-1}{\tan^2\theta_3+1}}.+\cos\theta_3^*=-\frac{\delta_{23}^*\tan^2\theta_3}{\tan^2\theta_3+1}\pm\sqrt{\left(\frac{\delta_{23}^*\tan^2\theta_3}{\tan^2\theta_3+1}\right)^2-\frac{\delta_{23}^{*2}\tan^2\theta_3-1}{\tan^2\theta_3+1}}.
 \end{align} \end{align}
  
行 514: 行 520:
 === Derivation of Quantity 12. Maximum K.E. transfer to a stationary particle === === Derivation of Quantity 12. Maximum K.E. transfer to a stationary particle ===
  
-** The first formula of the General Formulae **+** The General Formula **
  
 \begin{align} \begin{align}
行 589: 行 595:
 &= \frac{m_3\mp\sqrt{\frac{m_3m_4}{m_1}(m_3+m_4-m_1)}}{m_3+m_4}\boldsymbol{p}_1\\ &= \frac{m_3\mp\sqrt{\frac{m_3m_4}{m_1}(m_3+m_4-m_1)}}{m_3+m_4}\boldsymbol{p}_1\\
 \end{align} \end{align}
- 
  
 By using $T_4=\frac{|\boldsymbol{p}_4|^2}{2m_4}$ and $T_1=\frac{|\boldsymbol{p}_1|^2}{2m_1}$, By using $T_4=\frac{|\boldsymbol{p}_4|^2}{2m_4}$ and $T_1=\frac{|\boldsymbol{p}_1|^2}{2m_1}$,
行 626: 行 631:
 \end{align} \end{align}
  
-For the elastic scattering, $m_3 = m_1$ and $m_4 = m_2$. Therefore,+** The first formula for elastic scattering ** 
 + 
 +For elastic scattering, $m_3 = m_1$ and $m_4 = m_2$. Therefore,
 \begin{align} \begin{align}
 \boldsymbol{p}_4&=\frac{2m_2}{m_1+m_2}\boldsymbol{p}_1\\ \boldsymbol{p}_4&=\frac{2m_2}{m_1+m_2}\boldsymbol{p}_1\\
行 641: 行 648:
 \end{align} \end{align}
  
-** The second formula of the General Formulae **+ 
 + 
 +** 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, From the first formula,
行 658: 行 698:
 \end{align} \end{align}
  
-** Another derivation of the second formula of the General Formulae **+** Another derivation of the second formula for elastic scattering **
  
 If $m_2/m_1 \ll 1$, $|\boldsymbol{v}_\mathrm{cm}| \approx |\boldsymbol{v}_1|$. If $m_2/m_1 \ll 1$, $|\boldsymbol{v}_\mathrm{cm}| \approx |\boldsymbol{v}_1|$.
行 668: 行 708:
 \end{align} \end{align}
  
 +** The formula for N-N scattering **
 +
 +For N-N scattering, $m_1 = m_2 = m_\mathrm{N}$. Therefore,  from the formula above,
 +\begin{align}
 +T_\mathrm{max}
 +&=\frac{4m_1m_2}{(m_1+m_2)^2}T_1\\
 +&=\frac{4m_\mathrm{N}m_\mathrm{N}}{(m_\mathrm{N}+m_\mathrm{N})^2}T_1\\
 +&=\frac{4m_\mathrm{N}^2}{(2m_\mathrm{N})^2}T_1\\
 +&=T_1
 +\end{align}
  
 ==== General memo ==== ==== General memo ====
research/memos/kinematics/non-relativistic_kinematics.1506881325.txt.gz · 最終更新: 2017/10/02 03:08 by kobayash
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