From dedb366c3a78f61c64f6be627ea091e71e009f7d Mon Sep 17 00:00:00 2001 From: Prefetch Date: Sat, 2 Oct 2021 15:40:20 +0200 Subject: Expand knowledge base --- .../know/concept/rutherford-scattering/index.pdc | 242 +++++++++++++++++++++ .../concept/rutherford-scattering/one-body.png | Bin 0 -> 66663 bytes .../concept/rutherford-scattering/two-body.png | Bin 0 -> 40242 bytes 3 files changed, 242 insertions(+) create mode 100644 content/know/concept/rutherford-scattering/index.pdc create mode 100644 content/know/concept/rutherford-scattering/one-body.png create mode 100644 content/know/concept/rutherford-scattering/two-body.png (limited to 'content/know/concept/rutherford-scattering') diff --git a/content/know/concept/rutherford-scattering/index.pdc b/content/know/concept/rutherford-scattering/index.pdc new file mode 100644 index 0000000..481e4d1 --- /dev/null +++ b/content/know/concept/rutherford-scattering/index.pdc @@ -0,0 +1,242 @@ +--- +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: + + + + + +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: + + + + + +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. diff --git a/content/know/concept/rutherford-scattering/one-body.png b/content/know/concept/rutherford-scattering/one-body.png new file mode 100644 index 0000000..2f837bb Binary files /dev/null and b/content/know/concept/rutherford-scattering/one-body.png differ diff --git a/content/know/concept/rutherford-scattering/two-body.png b/content/know/concept/rutherford-scattering/two-body.png new file mode 100644 index 0000000..b67b6d1 Binary files /dev/null and b/content/know/concept/rutherford-scattering/two-body.png differ -- cgit v1.2.3