From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Thu, 20 Oct 2022 18:25:31 +0200 Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex' --- source/know/concept/rutherford-scattering/index.md | 86 +++++++++++----------- 1 file changed, 43 insertions(+), 43 deletions(-) (limited to 'source/know/concept/rutherford-scattering/index.md') diff --git a/source/know/concept/rutherford-scattering/index.md b/source/know/concept/rutherford-scattering/index.md index 7ce15ac..a7375d5 100644 --- a/source/know/concept/rutherford-scattering/index.md +++ b/source/know/concept/rutherford-scattering/index.md @@ -15,20 +15,20 @@ 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 $\theta$, +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 $b$. +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: +where $$r = |\vb{r}_1 - \vb{r}_2|$$ is the distance between 1 and 2: $$\begin{aligned} m_1 \dv{\vb{v}_1}{t} @@ -41,7 +41,7 @@ $$\begin{aligned} \end{aligned}$$ Using the [reduced mass](/know/concept/reduced-mass/) -$\mu \equiv m_1 m_2 / (m_1 \!+\! m_2)$, +$$\mu \equiv m_1 m_2 / (m_1 \!+\! m_2)$$, we turn this into a one-body problem: $$\begin{aligned} @@ -49,11 +49,11 @@ $$\begin{aligned} = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}}{r^3} \end{aligned}$$ -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. +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/) -$(r, \varphi, z)$: +$$(r, \varphi, z)$$: $$\begin{aligned} \vb{r} @@ -62,19 +62,19 @@ $$\begin{aligned} \end{aligned}$$ These new coordinates are sketched below, -where the origin represents $\vb{r}_1 = \vb{r}_2$. +where the origin represents $$\vb{r}_1 = \vb{r}_2$$. Crucially, note the symmetry: -if the "collision" occurs at $t = 0$, -then by comparing $t > 0$ and $t < 0$ -we can see that $v_x$ is unchanged for any given $\pm t$, -while $v_y$ simply changes sign: +if the "collision" occurs at $$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$$ simply changes sign: -From our expression for $\vb{r}$, -we can find $\vb{v}$ by differentiating with respect to time: +From our expression for $$\vb{r}$$, +we can find $$\vb{v}$$ by differentiating with respect to time: $$\begin{aligned} \vb{v} @@ -87,9 +87,9 @@ $$\begin{aligned} &= r' \:\vu{e}_r + r \varphi' \:\vu{e}_\varphi + z' \:\vu{e}_z \end{aligned}$$ -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: +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: $$\begin{aligned} \vb{r} @@ -99,8 +99,8 @@ $$\begin{aligned} = r' \:\vu{e}_r + r \varphi' \:\vu{e}_\varphi \end{aligned}$$ -Consequently, the angular momentum $\vb{L}$ is as follows, -pointing purely in the $z$-direction: +Consequently, the angular momentum $$\vb{L}$$ is as follows, +pointing purely in the $$z$$-direction: $$\begin{aligned} \vb{L}(t) @@ -110,8 +110,8 @@ $$\begin{aligned} \end{aligned}$$ 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: +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: $$\begin{aligned} \frac{b}{r(\pm \infty)} @@ -122,8 +122,8 @@ $$\begin{aligned} \end{aligned}$$ With this, we can rewrite -the magnitude of the angular momentum $\vb{L}$ as follows, -where the total velocity $|\vb{v}|$ is a constant, +the magnitude of the angular momentum $$\vb{L}$$ as follows, +where the total velocity $$|\vb{v}|$$ is a constant, thanks to conservation of energy: $$\begin{aligned} @@ -134,7 +134,7 @@ $$\begin{aligned} \end{aligned}$$ However, conveniently, -angular momentum is also conserved, i.e. $\vb{L}$ is constant in time: +angular momentum is also conserved, i.e. $$\vb{L}$$ is constant in time: $$\begin{aligned} \vb{L}'(t) @@ -144,12 +144,12 @@ $$\begin{aligned} = 0 \end{aligned}$$ -Where we have replaced $\mu \vb{v}'$ with the equation of motion. -Thanks to this, we can equate the two preceding expressions for $\vb{L}$, +Where we have replaced $$\mu \vb{v}'$$ with the equation of motion. +Thanks to this, we can equate the two preceding expressions for $$\vb{L}$$, leading to the relation below. Note the appearance of a new minus, -because the sketch shows that $\varphi' < 0$, -i.e. $\varphi$ decreases with increasing $t$: +because the sketch shows that $$\varphi' < 0$$, +i.e. $$\varphi$$ decreases with increasing $$t$$: $$\begin{aligned} - \mu r^2 \dv{\varphi}{t} @@ -160,7 +160,7 @@ $$\begin{aligned} \end{aligned}$$ Now, at last, we turn to the main equation of motion. -Its $y$-component is given by: +Its $$y$$-component is given by: $$\begin{aligned} \mu \dv{v_y}{t} @@ -170,8 +170,8 @@ $$\begin{aligned} = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{y}{r^3} \dd{t} \end{aligned}$$ -We replace $\dd{t}$ with our earlier relation, -and recognize geometrically that $y/r = \sin{\varphi}$: +We replace $$\dd{t}$$ with our earlier relation, +and recognize geometrically that $$y/r = \sin{\varphi}$$: $$\begin{aligned} \mu \dd{v_y} @@ -180,8 +180,8 @@ $$\begin{aligned} = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|} \dd{(\cos{\varphi})} \end{aligned}$$ -Integrating this from the initial state $i$ at $t = -\infty$ -to the final state $f$ at $t = \infty$ yields: +Integrating this from the initial state $$i$$ at $$t = -\infty$$ +to the final state $$f$$ at $$t = \infty$$ yields: $$\begin{aligned} \Delta v_y @@ -189,8 +189,8 @@ $$\begin{aligned} = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( \cos{\varphi_f} - \cos{\varphi_i} \big) \end{aligned}$$ -From symmetry, we see that $\varphi_i = \pi \!-\! \varphi_f$, -and that $\Delta v_y = v_{y,f} \!-\! v_{y,i} = 2 v_{y,f}$, such that: +From symmetry, we see that $$\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} @@ -198,8 +198,8 @@ $$\begin{aligned} = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( 2 \cos{\varphi_f} \big) \end{aligned}$$ -Furthermore, geometrically, at $t = \infty$ -we notice that $v_{y,f} = |\vb{v}| \sin{\varphi_f}$, +Furthermore, geometrically, at $$t = \infty$$ +we notice that $$v_{y,f} = |\vb{v}| \sin{\varphi_f}$$, leading to: $$\begin{aligned} @@ -208,7 +208,7 @@ $$\begin{aligned} \end{aligned}$$ Rearranging this yields the following equation -for the final polar angle $\varphi_f \equiv \varphi(\infty)$: +for the final polar angle $$\varphi_f \equiv \varphi(\infty)$$: $$\begin{aligned} \tan{\varphi_f} @@ -216,9 +216,9 @@ $$\begin{aligned} = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|^2 \mu} \end{aligned}$$ -However, we want $\theta$, not $\varphi_f$. +However, we want $$\theta$$, not $$\varphi_f$$. One last use of symmetry and geometry -tells us that $\theta = 2 \varphi_f$, +tells us that $$\theta = 2 \varphi_f$$, and we thus arrive at the celebrated **Rutherford scattering formula**: $$\begin{aligned} @@ -228,7 +228,7 @@ $$\begin{aligned} } \end{aligned}$$ -In fact, this formula is also valid if $q_1$ and $q_2$ have opposite signs; +In fact, this formula is also valid if $$q_1$$ and $$q_2$$ have opposite signs; in that case particle 2 is simply located on the other side of particle 1's trajectory. -- cgit v1.2.3