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--- research:memos:kinematics:non-relativistic_kinematics [2017/10/02 03:19] – kobayash
+++ research: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|}}
-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
@@ 行 14: / 行 14: @@
 \delta_{ij} = |\boldsymbol{v}_i|/|\boldsymbol{v}_j|,
 \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)**|
@@ 行 61: / 行 65: @@
 \tan\theta_3 = \frac{\sin\theta_3^*}{1+\cos\theta_3^*}\\
 \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}
 \tan\theta_{\rm cm} = \frac{\sin\theta_{\rm lab}}{\left(\cos\theta_{\rm lab}-v_{\rm cm}/|\boldsymbol{v}_3|\right)}
@@ 行 88: / 行 92: @@
 &\approx 2m_2\boldsymbol{v}_1^2
 \end{align}|\begin{align}
-T_\mathrm{max}=T1
+T_\mathrm{max}=T_1
-\end{align|
+\end{align}|
 === Derivation of Quantity 1. Total c.m. energy ===
@@ 行 355: / 行 359: @@
 is
 \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}
 Therefore,
 \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}
@@ 行 516: / 行 520: @@
 === Derivation of Quantity 12. Maximum K.E. transfer to a stationary particle ===
-** The General Formulae **
+** The General Formula **
 \begin{align}
@@ 行 591: / 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\\
 \end{align}
 By using $T_4=\frac{|\boldsymbol{p}_4|^2}{2m_4}$ and $T_1=\frac{|\boldsymbol{p}_1|^2}{2m_1}$,
@@ 行 643: / 行 646: @@
 \begin{align}
 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}