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| 両方とも前のリビジョン前のリビジョン次のリビジョン | 前のリビジョン | ||
| research:memos:kinematics:relativistic_kinematics [2020/07/30 00:18] – [References] kobayash | research:memos:kinematics:relativistic_kinematics [2022/01/09 12:16] (現在) – [TRIUMF Kinematics Handbook: Sec. V Relativistic Kinematics] kobayash | ||
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| 行 54: | 行 54: | ||
| | 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} |
| + | \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. $\theta_3^* = \theta_4^*$ | ||
| \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} |
| \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} |
| | 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" | ||
| + | \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 $\theta_4^* = \theta_3^*$, | By using $\theta_4^* = \theta_3^*$, | ||
| \begin{align} | \begin{align} | ||
| - | \tan\theta_4 = \frac{\sin\theta_3^*}{\gamma_2^*\left(|\boldsymbol{\beta}_2^*|/ | + | \tan\theta_4 |
| + | &= \frac{\sin\theta_3^*}{\gamma_2^*\left(|\boldsymbol{\beta}_2^*|/ | ||
| + | &= \frac{\sin\theta_3^*}{\gamma_2^*\left(\delta_{24}^*-\cos\theta_3^*\right)} | ||
| \end{align} | \end{align} | ||
| - | For elastic scattering, $m_2=m_4$, $|\boldsymbol{p}_2^*|=|\boldsymbol{p}_4^*|$, | + | For elastic scattering, $m_2=m_4$, $|\boldsymbol{p}_2^*|=|\boldsymbol{p}_4^*|$, |
| \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 1047: | 行 1061: | ||
| one can get | one can get | ||
| \begin{align} | \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 + |\boldsymbol{p}_3|^2, | + | |\boldsymbol{p}_4|^2 &= |\boldsymbol{p}_1 + \boldsymbol{p}_2|^2 - 2|\boldsymbol{p}_1 + \boldsymbol{p}_2||\boldsymbol{p}_3|\cos\theta_3 |
| E_4^2 &= (E_1 + E_2)^2-2(E_1+E_2)E_3 + E_3^2.\\ | E_4^2 &= (E_1 + E_2)^2-2(E_1+E_2)E_3 + E_3^2.\\ | ||
| \end{align} | \end{align} | ||
| 行 1057: | 行 1071: | ||
| the first equation becomes | the first equation becomes | ||
| \begin{align} | \begin{align} | ||
| - | E_4^2 -m_4^2 & | + | E_4^2 -m_4^2 & |
| \end{align} | \end{align} | ||
| By eliminating $E_4$ from the two equations, | By eliminating $E_4$ from the two equations, | ||
| \begin{align} | \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 + m_3^2\\ | + | 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 |
| \end{align} | \end{align} | ||
| Using the square of the invariant mass $W$ | Using the square of the invariant mass $W$ | ||
| 行 1068: | 行 1082: | ||
| \end{align} | \end{align} | ||
| \begin{align} | \begin{align} | ||
| - | & 2(E_1+E_2)E_3 - (W^2 + m_3^2-m_4^2) = 2\cos\theta|\boldsymbol{p}_1 + \boldsymbol{p}_2||\boldsymbol{p}_3|\\ | + | & 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|\boldsymbol{p}_1 + \boldsymbol{p}_2|\sqrt{E_3^2-m_3^2}\\ | + | \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|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\left[E_3^2-m_3^2\right]\\ | + | \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|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2 \right]E_3^2 | + | \Rightarrow & 4\left[(E_1+E_2)^2 - \cos^2\theta_3|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2 \right]E_3^2 |
| - | \Rightarrow & 4\left[1 - \cos^2\theta\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 | + | \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 |
| \end{align} | \end{align} | ||
| By using | By using | ||
| 行 1081: | 行 1095: | ||
| \end{align} | \end{align} | ||
| \begin{align} | \begin{align} | ||
| - | 4\left(1 - \beta_\mathrm{cm}^2\cos^2\theta\right)\left(\frac{E_3}{E_1+E_2}\right)^2 | + | 4\left(1 - \beta_\mathrm{cm}^2\cos^2\theta_3\right)\left(\frac{E_3}{E_1+E_2}\right)^2 |
| \end{align} | \end{align} | ||
| 行 1093: | 行 1107: | ||
| \begin{align} | \begin{align} | ||
| \frac{E_3}{E_1+E_2} | \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\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\left(\frac{m_3}{E_1+E_2}\right)^2\right]}}{2\left(1-\beta_\mathrm{cm}^2\cos^2\theta\right)}\\ | + | &= \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 | \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\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\left(\frac{m_3}{E_1+E_2}\right)^2\right]}}{2\left(1-\beta_\mathrm{cm}^2\cos^2\theta\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{\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\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\right)\beta_\mathrm{cm}^2\cos^2\theta\left(\frac{m_3}{E_1+E_2}\right)^2}}{2\left(1-\beta_\mathrm{cm}^2\cos^2\theta\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\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\right)\left(\frac{m_3}{E_1+E_2}\right)^2}}{2\left(1-\beta_\mathrm{cm}^2\cos^2\theta\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} | \end{align} | ||
| 行 1104: | 行 1118: | ||
| \begin{align} | \begin{align} | ||
| E_3 | E_3 | ||
| - | &= \frac{W^2 + m_3^2-m_4^2\pm\beta_\mathrm{cm}\cos\theta\sqrt{\left(W^2 +m_3^2-m_4^2\right)^2 -4m_3^2\left(W^2+\sin^2\theta|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right)}}{2\left(W^2+\sin^2\theta|\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)\\ |
| - | &= \frac{W^2 + m_3^2-m_4^2\pm\beta_\mathrm{cm}\cos\theta\sqrt{\left[W^2 -(m_3^2+m_4^2)\right]^2 -4m_3^2m_4^2 -4\sin^2\theta|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2}}{2\left(W^2+\sin^2\theta|\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} | \end{align} | ||
| - | When $\theta=0$ or $\pi$, | + | When $\theta_3=0$ or $\pi$, |
| \begin{align} | \begin{align} | ||
| E_3 | E_3 | ||
| 行 1142: | 行 1156: | ||
| Therefore, from | Therefore, from | ||
| \begin{align} | \begin{align} | ||
| - | 2(E_1+E_2)E_3 - (W^2 + m_3^2-m_4^2) = 2\cos\theta|\boldsymbol{p}_1 + \boldsymbol{p}_2|\sqrt{E_3^2-m_3^2}, | + | 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} | \end{align} | ||
| $E_3$ is | $E_3$ is | ||
| \begin{align} | \begin{align} | ||
| E_3 | E_3 | ||
| - | &= \frac{2(E_1+E_2)(W^2 + m_3^2-m_4^2){\pm}2\cos\theta|\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|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right]}}{4(E_1+E_2)^2-4\cos^2\theta|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2}\\ | + | &= \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|\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|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right]}}{2(E_1+E_2)^2-2\cos^2\theta|\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\sqrt{(W^2 + m_3^2-m_4^2)^2-4m_3^2\left[(E_1+E_2)^2-\cos^2\theta|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right]}}{2\left[(E_1+E_2)^2-\cos^2\theta|\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{(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\sqrt{\left(W^2 +m_3^2-m_4^2\right)^2 -4m_3^2\left(W^2+\sin^2\theta|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right)}}{2\left(W^2+\sin^2\theta|\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} | \end{align} | ||
| + | |||
| + | ==== $E_4$ from $\boldsymbol{p}_1$, | ||
| + | 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} | ||
| 行 1163: | 行 1223: | ||
| Similar with the derivation of $E_3$, | Similar with the derivation of $E_3$, | ||
| \begin{align} | \begin{align} | ||
| - | & 2(E_1+E_2)E_3 - (W^2 + m_3^2-m_4^2) = 2\cos\theta|\boldsymbol{p}_1 + \boldsymbol{p}_2||\boldsymbol{p}_3|\\ | + | & 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|\boldsymbol{p}_1 + \boldsymbol{p}_2||\boldsymbol{p}_3|+(W^2 + m_3^2-m_4^2).\\ | + | \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} | \end{align} | ||
| For simplicity, solve the following equation. | For simplicity, solve the following equation. | ||
| 行 1188: | 行 1248: | ||
| \begin{align} | \begin{align} | ||
| |\boldsymbol{p}_3| | |\boldsymbol{p}_3| | ||
| - | &= \frac{2\cos\theta|\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|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right]}}{4(E_1+E_2)^2-4\cos^2\theta|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2}\\ | + | &= \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|\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|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right]}}{2\left[(E_1+E_2)^2-\cos^2\theta|\boldsymbol{p}_1 + \boldsymbol{p}_2|^2\right]}\\ | + | &= \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} | \end{align} | ||
