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research:memos:kinematics:relativistic_kinematics [2022/01/04 15:28] – [$E_3$ from $\boldsymbol{p}_1$, $\boldsymbol{p}_2$, $E_1$ and $E_2$] kobayashresearch:memos:kinematics:relativistic_kinematics [2022/01/09 12:16] (現在) – [TRIUMF Kinematics Handbook: Sec. V Relativistic Kinematics] kobayash
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 | 7. c.m. to lab angle ($\theta_{\rm cm} \rightarrow \theta_{\rm lab}$) |\begin{align} | 7. c.m. to lab angle ($\theta_{\rm cm} \rightarrow \theta_{\rm lab}$) |\begin{align}
 \cos\theta_3 = \frac{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}{\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}} \cos\theta_3 = \frac{\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}|\begin{align}+\end{align} Another solution (not written in the document) \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} N.B. $\theta_3^* = \theta_4^*$ |\begin{align}
 \tan\theta_3 &= \frac{\sin\theta_3^*}{\gamma_2^*\left(\delta_{21}^*+\cos\theta_3^*\right)}\\ \tan\theta_3 &= \frac{\sin\theta_3^*}{\gamma_2^*\left(\delta_{21}^*+\cos\theta_3^*\right)}\\
 \tan\theta_4 &= \frac{1}{\gamma_2^*}\cot\frac{\theta_3^*}{2} \tan\theta_4 &= \frac{1}{\gamma_2^*}\cot\frac{\theta_3^*}{2}
-\end{align}|\begin{align}+\end{align} N.B. $\theta_3^* = \theta_4^*$|\begin{align}
 \tan\theta_3 = \frac{\sin\theta_3^*}{\gamma_2^*\left(1+\cos\theta_3^*\right)}\\ \tan\theta_3 = \frac{\sin\theta_3^*}{\gamma_2^*\left(1+\cos\theta_3^*\right)}\\
-\end{align}|+\end{align} N.B. $\theta_3^* = \theta_4^*$|
 | 8. lab to c.m. angle transformation ($\theta_\mathrm{lab} \rightarrow \theta_\mathrm{cm}$) |\begin{align} | 8. lab to c.m. angle transformation ($\theta_\mathrm{lab} \rightarrow \theta_\mathrm{cm}$) |\begin{align}
 \cos\theta_3^*=-\frac{\delta_{23}^*(\gamma_2^*\tan\theta_3)^2}{(\gamma_2^*\tan\theta_3)^2+1}\pm\sqrt{\left(\frac{\delta_{23}^*(\gamma_2^*\tan\theta_3)^2}{(\gamma_2^*\tan\theta_3)^2+1}\right)^2-\frac{\delta_{23}^{*2}(\gamma_2^*\tan\theta_3)^2-1}{(\gamma_2^*\tan\theta_3)^2+1}} \cos\theta_3^*=-\frac{\delta_{23}^*(\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}}
行 93: 行 96:
 \end{align}| \end{align}|
  
-=== Derivation of Quantity 1. Total c.m. energy ===+==== Derivation of Quantity 1. Total c.m. energy ====
 ** The General Formula ** ** The General Formula **
  
行 108: 行 111:
 \end{align} \end{align}
  
-=== Derivation of Quantity 2. c.m. momentum before the interaction ===+==== Derivation of Quantity 2. c.m. momentum before the interaction ====
 ** The General Formula ** ** The General Formula **
  
行 164: 行 167:
 \end{align} \end{align}
  
-=== Derivation of Quantity 3. c.m. momentum after the interaction ===+==== Derivation of Quantity 3. c.m. momentum after the interaction ====
 ** The General Formula ** ** The General Formula **
  
行 187: 行 190:
 \end{align} \end{align}
  
-=== Derivation of Quantity 4. Velocity of the c.m. ===+==== Derivation of Quantity 4. Velocity of the c.m. ====
 ** The General Formula ** ** The General Formula **
  
行 207: 行 210:
 \end{align} \end{align}
  
-=== Derivation of Quantity 5. $\gamma$ of the c.m. ===+==== Derivation of Quantity 5. $\gamma$ of the c.m. ====
 ** The General Formula ** ** The General Formula **
  
行 232: 行 235:
  
  
-=== Derivation of Quantity 6. Maximum lab scattering angle ===+==== Derivation of Quantity 6. Maximum lab scattering angle ====
 ** The General Formula ** ** The General Formula **
  
行 289: 行 292:
 Therefore, $\tan\theta_3$ becomes infinite at $\cos\theta_3^*=-1$. In that case, $\mathrm{3max}=90^\circ$. Therefore, $\tan\theta_3$ becomes infinite at $\cos\theta_3^*=-1$. In that case, $\mathrm{3max}=90^\circ$.
  
-=== Derivation of Quantity 7. c.m. to lab angle ($\theta_{\rm cm} \rightarrow \theta_{\rm lab}$) ===+==== Derivation of Quantity 7. c.m. to lab angle ($\theta_{\rm cm} \rightarrow \theta_{\rm lab}$) ====
 ** The General Formula ** ** The General Formula **
  
行 334: 行 337:
 &= \frac{\gamma_2^*\left(\delta_{23}^*+\cos\theta_3^*\right)}{\sqrt{\sin^2\theta_3^*+\gamma_2^{*2}\left(\delta_{23}^*+\cos\theta_3^*\right)^2}} &= \frac{\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} \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 ** ** The first formula for Elastic Scattering **
行 382: 行 394:
 By using $\theta_4^* = \theta_3^*$, $\beta_{\rm cm} = |\boldsymbol{\beta}_2^*|$ and $\gamma_{\rm cm} = \gamma_2^*$, By using $\theta_4^* = \theta_3^*$, $\beta_{\rm cm} = |\boldsymbol{\beta}_2^*|$ and $\gamma_{\rm cm} = \gamma_2^*$,
 \begin{align} \begin{align}
-\tan\theta_4 = \frac{\sin\theta_3^*}{\gamma_2^*\left(|\boldsymbol{\beta}_2^*|/|\boldsymbol{\beta}_4^*|-\cos\theta_3^*\right)}+\tan\theta_4 
 +&= \frac{\sin\theta_3^*}{\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} \end{align}
  
-For elastic scattering, $m_2=m_4$, $|\boldsymbol{p}_2^*|=|\boldsymbol{p}_4^*|$, and $|\boldsymbol{\beta}_2^*|=|\boldsymbol{\beta}_4^*|$. Therefore,+For elastic scattering, $m_2=m_4$, $|\boldsymbol{p}_2^*|=|\boldsymbol{p}_4^*|$, $|\boldsymbol{\beta}_2^*|=|\boldsymbol{\beta}_4^*|$, and $\delta_{24}^*=1$. Therefore,
 \begin{align} \begin{align}
 \tan\theta_4 = \frac{\sin\theta_3^*}{\gamma_2^*(1-\cos\theta_3^*)} \tan\theta_4 = \frac{\sin\theta_3^*}{\gamma_2^*(1-\cos\theta_3^*)}
行 407: 行 421:
 \end{align} \end{align}
  
-=== Derivation of Quantity 8. lab to c.m. angle transformation ($\theta_\mathrm{cm} \rightarrow \theta_\mathrm{lab}$) ===+==== Derivation of Quantity 8. lab to c.m. angle transformation ($\theta_\mathrm{cm} \rightarrow \theta_\mathrm{lab}$) ====
 ** The General Formula ** ** The General Formula **
  
行 466: 行 480:
 \end{align} \end{align}
  
-=== Derivation of Quantity 9. Solid angle transformation (Jacobian) ===+==== Derivation of Quantity 9. Solid angle transformation (Jacobian) ====
  
 ** The first formula of the General Formulae ** ** The first formula of the General Formulae **
行 526: 行 540:
 \end{align} \end{align}
  
-===  Derivation of Quantity 10. Relations between the $\gamma$ factors ===+====  Derivation of Quantity 10. Relations between the $\gamma$ factors ====
  
 ** The first formula of the General Formulae ** ** The first formula of the General Formulae **
行 620: 行 634:
  
  
-=== Derivation of Quantity 11. Lab quantity relations ===+==== Derivation of Quantity 11. Lab quantity relations ====
  
 ** The first formula of the General Formulae ** ** The first formula of the General Formulae **
行 711: 行 725:
  
  
-=== Derivation of Quantity 12. Maximum K.E. transfer to a stationary particle ===+==== Derivation of Quantity 12. Maximum K.E. transfer to a stationary particle ====
  
  
行 1174: 行 1188:
  
 ==== $E_4$ from $\boldsymbol{p}_1$, $\boldsymbol{p}_2$, $E_1$ and $E_2$ ==== ==== $E_4$ from $\boldsymbol{p}_1$, $\boldsymbol{p}_2$, $E_1$ and $E_2$ ====
 +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 the particle is swapped (elastic scattering and $m_3=m_2$, $m_4=m_1$),+If $|\boldsymbol{p}_2| = 0$$E_2 = m_2$ and elastic scattering ($m_3=m_1$, $m_4=m_2$),
 \begin{align} \begin{align}
-E_3 +E_4 
-&= \frac{(E_1+m_2)(W^2 + m_2^2-m_1^2){\pm}\cos\theta|\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|\boldsymbol{p}_1|^2\right]}}{2(E_1+m_2)^2-2\cos^2\theta|\boldsymbol{p}_1|^2}\\ +&= \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|\boldsymbol{p}_1|^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|\boldsymbol{p}_1|\sqrt{(2m_2^2+2E_1m_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|\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|\boldsymbol{p}_1|\sqrt{(m_2^2+E_1m_2)^2-m_2^2\left[(E_1+m_2)^2-\cos^2\theta|\boldsymbol{p}_1|^2\right]}}{(E_1+m_2)^2-\cos^2\theta|\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|\boldsymbol{p}_1|\sqrt{(m_2+E_1)^2-\left[(E_1+m_2)^2-\cos^2\theta|\boldsymbol{p}_1|^2\right]}}{(E_1+m_2)^2-\cos^2\theta|\boldsymbol{p}_1|^2}m_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|\boldsymbol{p}_1|^2}{(E_1+m_2)^2-\cos^2\theta|\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} \end{align}
-$E_3 \ne m2$, and then+$E_4 \ne m_2$, and then
 \begin{align} \begin{align}
-E_3 +E_4 
-&= \frac{(E_1+m_2)^2+\cos^2\theta|\boldsymbol{p}_1|^2}{(E_1+m_2)^2-\cos^2\theta|\boldsymbol{p}_1|^2}m_2\\+&= \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} \end{align}
  
research/memos/kinematics/relativistic_kinematics.1641277694.txt.gz · 最終更新: 2022/01/04 15:28 by kobayash
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