+
+
+
+
+We introduce some new terms, and reorganize the expression:
+
+$$\begin{aligned}
+ 2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
+ &= 2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
+ + \gamma_g \rho_{ee} - \gamma_g \rho_{ee}
+ + \gamma_e \rho_{gg} - \gamma_e \rho_{gg}
+ \\
+ &= \gamma_g (\rho_{gg} + \rho_{ee}) - \gamma_e (\rho_{gg} + \rho_{ee})
+ + \gamma_g (\rho_{gg} - \rho_{ee}) + \gamma_e (\rho_{gg} - \rho_{ee})
+\end{aligned}$$
+
+Since the total probability $\rho_{gg} + \rho_{ee} = 1$,
+and $d \equiv \rho_{ee} - \rho_{gg}$, this reduces to:
+
+$$\begin{aligned}
+ 2 \gamma_g \rho_{gg} - 2 \gamma_e \rho_{ee}
+ &= \gamma_g - \gamma_e - (\gamma_g + \gamma_e) d
+ \\
+ &= (\gamma_g + \gamma_e) \Big( \frac{\gamma_g - \gamma_e}{\gamma_g + \gamma_e} - d \Big)
+ \\
+ &= \gamma_\parallel ( d_0 - d )
+\end{aligned}$$
+
+
+
+
+
+From our expression for $\vb{r}$,
+we can find $\vb{v}$ by differentiating with respect to time:
+
+$$\begin{aligned}
+ \vb{v}
+ &= \big( r' \cos{\varphi} - r \varphi' \sin{\varphi} \big) \:\vu{e}_x
+ + \big( r' \sin{\varphi} + r \varphi' \cos{\varphi} \big) \:\vu{e}_y + z' \:\vu{e}_z
+ \\
+ &= r' \: \big( \cos{\varphi} \:\vu{e}_x + \sin{\varphi} \:\vu{e}_y \big)
+ + r \varphi' \: \big( \!-\! \sin{\varphi} \:\vu{e}_x + \cos{\varphi} \:\vu{e}_y \big) + z' \:\vu{e}_z
+ \\
+ &= r' \:\vu{e}_r + r \varphi' \:\vu{e}_\varphi + z' \:\vu{e}_z
+\end{aligned}$$
+
+Where we have recognized the basis vectors $\vu{e}_r$ and $\vu{e}_\varphi$.
+If we choose the coordinate system such that all dynamics are in the $(x,y)$-plane,
+i.e. $z(t) = 0$, we have:
+
+$$\begin{aligned}
+ \vb{r}
+ = r \: \vu{e}_r
+ \qquad \qquad
+ \vb{v}
+ = r' \:\vu{e}_r + r \varphi' \:\vu{e}_\varphi
+\end{aligned}$$
+
+Consequently, the angular momentum $\vb{L}$ is as follows,
+pointing purely in the $z$-direction:
+
+$$\begin{aligned}
+ \vb{L}(t)
+ = \mu \vb{r} \cross \vb{v}
+ = \mu \big( r \vu{e}_r \cross r \varphi' \vu{e}_\varphi \big)
+ = \mu r^2 \varphi' \:\vu{e}_z
+\end{aligned}$$
+
+Now, from the figure above,
+we can argue geometrically that at infinity $t = \pm \infty$,
+the ratio $b/r$ is related to the angle $\chi$ between $\vb{v}$ and $\vb{r}$ like so:
+
+$$\begin{aligned}
+ \frac{b}{r(\pm \infty)}
+ = \sin{\chi(\pm \infty)}
+ \qquad \quad
+ \chi(t)
+ \equiv \measuredangle(\vb{r}, \vb{v})
+\end{aligned}$$
+
+With this, we can rewrite
+the magnitude of the angular momentum $\vb{L}$ as follows,
+where the total velocity $|\vb{v}|$ is a constant,
+thanks to conservation of energy:
+
+$$\begin{aligned}
+ \big| \vb{L}(\pm \infty) \big|
+ = \mu \big| \vb{r} \cross \vb{v} \big|
+ = \mu r |\vb{v}| \sin{\chi}
+ = \mu b |\vb{v}|
+\end{aligned}$$
+
+However, conveniently,
+angular momentum is also conserved, i.e. $\vb{L}$ is constant in time:
+
+$$\begin{aligned}
+ \vb{L}'(t)
+ &= \mu \big( \vb{r} \cross \vb{v}' + \vb{v} \cross \vb{v} \big)
+ = \vb{r} \cross (\mu \vb{v}')
+ = \vb{r} \cross \Big( \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}}{r^3} \Big)
+ = 0
+\end{aligned}$$
+
+Where we have replaced $\mu \vb{v}'$ with the equation of motion.
+Thanks to this, we can equate the two preceding expressions for $\vb{L}$,
+leading to the relation below.
+Note the appearance of a new minus,
+because the sketch shows that $\varphi' < 0$,
+i.e. $\varphi$ decreases with increasing $t$:
+
+$$\begin{aligned}
+ - \mu r^2 \dv{\varphi}{t}
+ = \mu b |\vb{v}|
+ \quad \implies \quad
+ \dd{t}
+ = - \frac{r^2}{b |\vb{v}|} \dd{\varphi}
+\end{aligned}$$
+
+Now, at last, we turn to the main equation of motion.
+Its $y$-component is given by:
+
+$$\begin{aligned}
+ \mu \dv{v_y}{t}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{y}{r^3}
+ \quad \implies \quad
+ \mu \dd{v_y}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{y}{r^3} \dd{t}
+\end{aligned}$$
+
+We replace $\dd{t}$ with our earlier relation,
+and recognize geometrically that $y/r = \sin{\varphi}$:
+
+$$\begin{aligned}
+ \mu \dd{v_y}
+ = - \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|} \frac{y}{r} \dd{\varphi}
+ = - \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|} \sin{\varphi} \dd{\varphi}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|} \dd{(\cos{\varphi})}
+\end{aligned}$$
+
+Integrating this from the initial state $i$ at $t = -\infty$
+to the final state $f$ at $t = \infty$ yields:
+
+$$\begin{aligned}
+ \Delta v_y
+ \equiv \int_{i}^{f} \dd{v_y}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( \cos{\varphi_f} - \cos{\varphi_i} \big)
+\end{aligned}$$
+
+From symmetry, we see that $\varphi_i = \pi \!-\! \varphi_f$,
+and that $\Delta v_y = v_{y,f} \!-\! v_{y,i} = 2 v_{y,f}$, such that:
+
+$$\begin{aligned}
+ 2 v_{y,f}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( \cos{\varphi_f} - \cos\!(\pi \!-\! \varphi_f) \big)
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( 2 \cos{\varphi_f} \big)
+\end{aligned}$$
+
+Furthermore, geometrically, at $t = \infty$
+we notice that $v_{y,f} = |\vb{v}| \sin{\varphi_f}$,
+leading to:
+
+$$\begin{aligned}
+ 2 |\vb{v}| \sin{\varphi_f}
+ = \frac{q_1 q_2}{2 \pi \varepsilon_0 b |\vb{v}| \mu} \cos{\varphi_f}
+\end{aligned}$$
+
+Rearranging this yields the following equation
+for the final polar angle $\varphi_f \equiv \varphi(\infty)$:
+
+$$\begin{aligned}
+ \tan{\varphi_f}
+ = \frac{\sin{\varphi_f}}{\cos{\varphi_f}}
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|^2 \mu}
+\end{aligned}$$
+
+However, we want $\theta$, not $\varphi_f$.
+One last use of symmetry and geometry
+tells us that $\theta = 2 \varphi_f$,
+and we thus arrive at the celebrated **Rutherford scattering formula**:
+
+$$\begin{aligned}
+ \boxed{
+ \tan\!\Big( \frac{\theta}{2} \Big)
+ = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|^2 \mu}
+ }
+\end{aligned}$$
+
+
+
+## References
+1. P.M. Bellan,
+ *Fundamentals of plasma physics*,
+ 1st edition, Cambridge.
+2. M. Salewski, A.H. Nielsen,
+ *Plasma physics: lecture notes*,
+ 2021, unpublished.
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