Improve knowledge base

1 files changed, 37 insertions, 33 deletions

diff --git a/source/know/concept/rutherford-scattering/index.md b/source/know/concept/rutherford-scattering/index.md
index edf391c..7a2a1f2 100644
--- a/source/know/concept/rutherford-scattering/index.md
+++ b/source/know/concept/rutherford-scattering/index.md
@@ -27,13 +27,13 @@ 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 $$r \equiv |\vb{r}_1 \!-\! \vb{r}_2|$$ is the distance between 1 and 2:
 
 $$\begin{aligned}
     m_1 \dv{\vb{v}_1}{t}
     = \vb{F}_1
     = \frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{\vb{r}_1 - \vb{r}_2}{r^3}
-    \qquad \quad
+    \qquad \qquad
     m_2 \dv{\vb{v}_2}{t}
     = \vb{F}_2
     = - \vb{F}_1
@@ -56,8 +56,9 @@ $$(r, \varphi, z)$$:
 
 $$\begin{aligned}
     \vb{r}
-    = r \cos{\varphi} \:\vu{e}_x + r \sin{\varphi} \:\vu{e}_y + z \:\vu{e}_z
-    = r \:\vu{e}_r + z \:\vu{e}_z
+    &= r \cos{\varphi} \:\vu{e}_x + r \sin{\varphi} \:\vu{e}_y + z \:\vu{e}_z
+    \\
+    &= r \:\vu{e}_r + z \:\vu{e}_z
 \end{aligned}$$
 
 These new coordinates are sketched below,
@@ -76,6 +77,7 @@ we can find $$\vb{v}$$ by differentiating with respect to time:
 
 $$\begin{aligned}
     \vb{v}
+    = \vb{r}'
     &= \big( r' \cos{\varphi} - r \varphi' \sin{\varphi} \big) \:\vu{e}_x
     + \big( r' \sin{\varphi} + r \varphi' \cos{\varphi} \big) \:\vu{e}_y + z' \:\vu{e}_z
     \\
@@ -107,32 +109,34 @@ $$\begin{aligned}
     = \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:
+Now, in the figure above, imagine a right-angled triangle
+with hypotenuse $$\vb{r}$$ and short side $$b$$.
+When $$t \to +\infty$$, trigonometry tells us the following,
+where $$\chi$$ is the final angle between $$\vb{v}$$ and $$\vb{r}$$:
 
 $$\begin{aligned}
-    \frac{b}{r(\pm \infty)}
-    = \sin{\chi(\pm \infty)}
-    \qquad \quad
-    \chi(t)
-    \equiv \measuredangle(\vb{r}, \vb{v})
+    \lim_{t \to +\infty} \frac{b}{r(t)}
+    = \sin{\chi}
+    \qquad \qquad
+    \chi
+    \equiv
+    \lim_{t \to +\infty} \measuredangle(\vb{r}(t), \vb{v}(t))
 \end{aligned}$$
 
-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:
+With this, we can rewrite the magnitude of the angular momentum $$\vb{L}$$ as follows,
+where the relative speed $$|\vb{v}|$$ is a constant thanks to energy conservation:
 
 $$\begin{aligned}
-    \big| \vb{L}(\pm \infty) \big|
-    = \mu \big| \vb{r} \cross \vb{v} \big|
+    \lim_{t \to +\infty}
+    \big| \vb{L}(t) \big|
     = \mu r |\vb{v}| \sin{\chi}
     = \mu b |\vb{v}|
 \end{aligned}$$
 
-However, conveniently,
-angular momentum is also conserved, i.e. $$\vb{L}$$ is constant in time:
+This is useful, because angular momentum is conserved,
+i.e. $$\vb{L}$$ is constant in time.
+We prove this by using the product rule of differentiation,
+and replacing $$\mu \vb{v}'$$ with the reduced equation of motion:
 
 $$\begin{aligned}
     \vb{L}'(t)
@@ -142,8 +146,8 @@ $$\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 $$\vb{L}$$,
+Thanks to this, we can equate the two preceding expressions
+for the magnitude $$|\vb{L}|$$,
 leading to the relation below.
 Note the appearance of a new minus,
 because the sketch shows that $$\varphi' < 0$$,
@@ -178,8 +182,8 @@ $$\begin{aligned}
     = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|} \dd{(\cos{\varphi})}
 \end{aligned}$$
 
-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 $$t \to -\infty$$
+to the final state $$f$$ at $$t \to +\infty$$ yields:
 
 $$\begin{aligned}
     \Delta v_y
@@ -187,18 +191,18 @@ $$\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 $$\Delta v_y = v_{y,f} \!-\! v_{y,i} = 2 v_{y,f}$$,
+and that $$\varphi_i = \pi \!-\! \varphi_f$$, such that:
 
 $$\begin{aligned}
-    2 v_{y,f}
+    \Delta v_y
+    = 2 v_{y,f}
     = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( \cos{\varphi_f} - \cos(\pi \!-\! \varphi_f) \big)
     = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}| \mu} \big( 2 \cos{\varphi_f} \big)
 \end{aligned}$$
 
-Furthermore, geometrically, at $$t = \infty$$
-we notice that $$v_{y,f} = |\vb{v}| \sin{\varphi_f}$$,
-leading to:
+Furthermore, geometrically for $$t \to +\infty$$
+we notice that $$v_{y,f} = |\vb{v}| \sin{\varphi_f}$$, leading to:
 
 $$\begin{aligned}
     2 |\vb{v}| \sin{\varphi_f}
@@ -206,7 +210,7 @@ $$\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 $$\varphi_f$$:
 
 $$\begin{aligned}
     \tan{\varphi_f}
@@ -214,14 +218,14 @@ $$\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 the deflection angle $$\theta$$, 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**:
 
 $$\begin{aligned}
     \boxed{
-        \tan\!\Big( \frac{\theta}{2} \Big)
+        \tan\!\bigg( \frac{\theta}{2} \bigg)
         = \frac{q_1 q_2}{4 \pi \varepsilon_0 b |\vb{v}|^2 \mu}
     }
 \end{aligned}$$