Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

1 files changed, 43 insertions, 43 deletions

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:
 
 <a href="two-body.png">
 <img src="two-body.png" style="width:50%">
 </a>
 
-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:
 
 <a href="one-body.png">
 <img src="one-body.png" style="width:60%">
 </a>
 
-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.