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+---
+title: "Rutherford scattering"
+firstLetter: "R"
+publishDate: 2021-10-02
+categories:
+- Physics
+
+date: 2021-09-23T16:22:07+02:00
+draft: false
+markup: pandoc
+---
+
+# Rutherford scattering
+
+**Rutherford scattering** or **Coulomb scattering**
+is an elastic pseudo-collision of two electrically charged particles.
+It is not a true collision, and is caused by Coulomb repulsion.
+
+The general idea is illustrated below.
+Consider two particles 1 and 2, with the same charge sign.
+Let 2 be initially at rest, and 1 approach it with velocity $\vb{v}_1$.
+Coulomb repulsion causes 1 to deflect by an angle $\theta$,
+and pushes 2 away in the process:
+
+<a href="two-body.png">
+<img src="two-body.png" style="width:50%;display:block;margin:auto;">
+</a>
+
+Here, $b$ is called the **impact parameter**.
+Intuitively, we expect $\theta$ to be larger for smaller $b$.
+
+By combining Coulomb's law with Newton's laws,
+these particles' equations of motion are found to be as follows,
+where $r = |\vb{r}_1 - \vb{r}_2|$ is the distance between 1 and 2:
+
+$$\begin{aligned}
+    m_1 \dv{\vb{v}_1}{t}
+    = \vb{F}_1
+    = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}_1 - \vb{r}_2}{r^3}
+    \qquad \quad
+    m_2 \dv{\vb{v}_2}{t}
+    = \vb{F}_2
+    = - \vb{F}_1
+\end{aligned}$$
+
+Using the [reduced mass](/know/concept/reduced-mass/)
+$\mu \equiv m_1 m_2 / (m_1 \!+\! m_2)$,
+we turn this into a one-body problem:
+
+$$\begin{aligned}
+    \mu \dv{\vb{v}}{t}
+    = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}}{r^3}
+\end{aligned}$$
+
+Where $\vb{v} \equiv \vb{v}_1 \!-\! \vb{v}_2$ is the relative velocity,
+and $\vb{r} \equiv \vb{r}_1 \!-\! \vb{r}_2$ is the relative position.
+The latter is as follows in
+[cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/)
+$(r, \varphi, z)$:
+
+$$\begin{aligned}
+    \vb{r}
+    = r \cos{\varphi} \:\vu{e}_x + r \sin{\varphi} \:\vu{e}_y + z \:\vu{e}_z
+    = r \:\vu{e}_r + z \:\vu{e}_z
+\end{aligned}$$
+
+These new coordinates are sketched below,
+where the origin represents $\vb{r}_1 = \vb{r}_2$.
+Crucially, note the symmetry:
+if the "collision" occurs at $t = 0$,
+then by comparing $t > 0$ and $t < 0$
+we can see that $v_x$ is unchanged for any given $\pm t$,
+while $v_y$ simply changes sign:
+
+<a href="one-body.png">
+<img src="one-body.png" style="width:60%;display:block;margin:auto;">
+</a>
+
+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.