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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:non-relativistic_kinematics [2017/10/02 21:05] – kobayash | research:memos:kinematics:non-relativistic_kinematics [2020/07/29 17:10] (現在) – [Formulae for Non-relativistic Kinematics] kobayash | ||
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| 行 65: | 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}/ | \tan\theta_{\rm cm} = \frac{\sin\theta_{\rm lab}}{\left(\cos\theta_{\rm lab}-v_{\rm cm}/ | ||
| 行 359: | 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 595: | 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 647: | 行 646: | ||
| \begin{align} | \begin{align} | ||
| T_\mathrm{max}=T_4=\frac{4m_1m_2}{(m_1+m_2)^2}T_1 | T_\mathrm{max}=T_4=\frac{4m_1m_2}{(m_1+m_2)^2}T_1 | ||
| + | \end{align} | ||
| + | |||
| + | |||
| + | |||
| + | ** Another derivation of the first formula for elastic scattering ** | ||
| + | |||
| + | From the above discussion, $|\boldsymbol{v}_4|$ becomes maximum at $\theta_4^* = \theta_\mathrm{cm}=180^\circ$. In this case, $\theta_4 = 0^\circ$ and $\boldsymbol{v}_4$ has the same direction as $\boldsymbol{v}_1$. For $\theta_4 = 0^\circ$, from the law of conservation of energy and momentum, | ||
| + | |||
| + | \begin{align} | ||
| + | \frac{1}{2}m_1\boldsymbol{v}_1^2 = \frac{1}{2}m_1\boldsymbol{v}_3^2 + \frac{1}{2}m_2\boldsymbol{v}_4^2\\ | ||
| + | m_1\boldsymbol{v}_1 = m_1\boldsymbol{v}_3 + m_2\boldsymbol{v}_4\\ | ||
| + | \end{align} | ||
| + | |||
| + | From the second equation, | ||
| + | \begin{align} | ||
| + | \boldsymbol{v}_3 = \frac{m_1\boldsymbol{v}_1 - m_2\boldsymbol{v}_4}{m_1}. | ||
| + | \end{align} | ||
| + | |||
| + | By substituting this formula into the formula of the law of conservation of energy, | ||
| + | |||
| + | \begin{align} | ||
| + | \frac{1}{2}m_1\boldsymbol{v}_1^2 | ||
| + | &= \frac{1}{2}m_1\frac{(m_1\boldsymbol{v}_1 - m_2\boldsymbol{v}_4)^2}{m_1^2} + \frac{1}{2}m_2\boldsymbol{v}_4^2\\ | ||
| + | &= \frac{1}{2}m_1\frac{m_1^2\boldsymbol{v}_1^2 + m_2^2\boldsymbol{v}_4^2 - 2m_1m_2|\boldsymbol{v}_1||\boldsymbol{v}_4|}{m_1^2} + \frac{1}{2}m_2\boldsymbol{v}_4^2\\ | ||
| + | \Rightarrow 0 &= \frac{1}{2}m_1\frac{m_2^2\boldsymbol{v}_4^2 - 2m_1m_2|\boldsymbol{v}_1||\boldsymbol{v}_4|}{m_1^2} + \frac{1}{2}m_2\boldsymbol{v}_4^2 \\ | ||
| + | \Rightarrow 0 &= \frac{m_2|\boldsymbol{v}_4| - 2m_1|\boldsymbol{v}_1|}{m_1} + |\boldsymbol{v}_4| \\ | ||
| + | \Rightarrow 2|\boldsymbol{v}_1| &= \left(\frac{m_2}{m_1}+1\right)|\boldsymbol{v}_4|\\ | ||
| + | \Rightarrow |\boldsymbol{v}_4| | ||
| + | &= 2\frac{1}{\frac{m_2}{m_1}+1}|\boldsymbol{v}_1|\\ | ||
| + | &= \frac{2m_1}{m_1+m_2}|\boldsymbol{v}_1|\\ | ||
| + | \Rightarrow \boldsymbol{v}_4^2 &= \frac{4m_1^2}{(m_1+m_2)^2}\boldsymbol{v}_1^2\\ | ||
| + | \Rightarrow \frac{1}{2}m_4\boldsymbol{v}_4^2 &= \frac{4m_4m_1}{(m_1+m_2)^2}\frac{1}{2}m_1\boldsymbol{v}_1^2\\ | ||
| + | \Rightarrow T_\mathrm{max} &= T_4 = \frac{4m_4m_1}{(m_1+m_2)^2}T_1\\ | ||
| \end{align} | \end{align} | ||
