Introduce collapsible proofs to some articles

1 files changed, 95 insertions, 59 deletions

diff --git a/content/know/concept/curvilinear-coordinates/index.pdc b/content/know/concept/curvilinear-coordinates/index.pdc
index e1c0465..925eda3 100644
--- a/content/know/concept/curvilinear-coordinates/index.pdc
+++ b/content/know/concept/curvilinear-coordinates/index.pdc
@@ -50,7 +50,7 @@ and [parabolic cylindrical coordinates](/know/concept/parabolic-cylindrical-coor
 
 In the following subsections,
 we derive general formulae to convert expressions
-from Cartesian coordinates in the new orthogonal system $(x_1, x_2, x_3)$.
+from Cartesian coordinates to the new orthogonal system $(x_1, x_2, x_3)$.
 
 
 ## Basis vectors
@@ -93,7 +93,26 @@ $$\begin{aligned}
 
 ## Gradient
 
-For a given direction $\dd{\ell}$, we know that
+In an orthogonal coordinate system,
+the gradient $\nabla f$ of a scalar $f$ is as follows,
+where $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$
+are the basis unit vectors respectively corresponding to $x_1$, $x_2$ and $x_3$:
+
+$$\begin{gathered}
+    \boxed{
+        \nabla f
+        = \vu{e}_1 \frac{1}{h_1} \pdv{f}{x_1}
+        + \vu{e}_2 \frac{1}{h_2} \pdv{f}{x_2}
+        + \vu{e}_3 \frac{1}{h_3} \pdv{f}{x_3}
+    }
+\end{gathered}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-grad"/>
+<label for="proof-grad">Proof</label>
+<div class="hidden">
+<label for="proof-grad">Proof.</label>
+For a direction $\dd{\ell}$, we know that
 $\dv*{f}{\ell}$ is the component of $\nabla f$ in that direction:
 
 $$\begin{aligned}
@@ -104,7 +123,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Where $\vu{u}$ is simply a unit vector in the direction of $\dd{\ell}$.
-We can thus find an expression for the gradient $\nabla f$
+We thus find the expression for the gradient $\nabla f$
 by choosing $\dd{\ell}$ to be $h_1 \dd{x_1}$, $h_2 \dd{x_2}$ and $h_3 \dd{x_3}$ in turn:
 
 $$\begin{gathered}
@@ -112,49 +131,59 @@ $$\begin{gathered}
     = \vu{e}_1 \dv{x_1}{\ell} \pdv{f}{x_1}
     + \vu{e}_2 \dv{x_2}{\ell} \pdv{f}{x_2}
     + \vu{e}_3 \dv{x_3}{\ell} \pdv{f}{x_3}
-    \\
-    \boxed{
-        \nabla f
-        = \vu{e}_1 \frac{1}{h_1} \pdv{f}{x_1}
-        + \vu{e}_2 \frac{1}{h_2} \pdv{f}{x_2}
-        + \vu{e}_3 \frac{1}{h_3} \pdv{f}{x_3}
-    }
 \end{gathered}$$
-
-Where $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$
-are the basis unit vectors respectively corresponding to $x_1$, $x_2$ and $x_3$.
+</div>
+</div>
 
 
 ## Divergence
 
-Consider a vector $\vb{V}$ in the target coordinate system
-with components $V_1$, $V_2$ and $V_3$:
+The divergence of a vector $\vb{V} = \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3$
+in an orthogonal system is given by:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{V}
+        = \frac{1}{h_1 h_2 h_3}
+        \Big( \pdv{(h_2 h_3 V_1)}{x_1} + \pdv{(h_1 h_3 V_2)}{x_2} + \pdv{(h_1 h_2 V_3)}{x_3} \Big)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-div"/>
+<label for="proof-div">Proof</label>
+<div class="hidden">
+<label for="proof-div">Proof.</label>
+As preparation, we rewrite $\vb{V}$ as follows
+to introduce the scale factors:
 
 $$\begin{aligned}
     \vb{V}
-    &= \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3
-    \\
     &= \vu{e}_1 \frac{1}{h_2 h_3} (h_2 h_3 V_1)
     + \vu{e}_2 \frac{1}{h_1 h_3} (h_1 h_3 V_2)
     + \vu{e}_3 \frac{1}{h_1 h_2} (h_1 h_2 V_3)
 \end{aligned}$$
 
-We take only the $\vu{e}_1$-component of this vector,
-and expand its divergence using a vector identity,
-where $f = h_2 h_3 V_1$ is a scalar
-and $\vb{U} = \vu{e}_1 / (h_2 h_3)$ is a vector:
+We start by taking only the $\vu{e}_1$-component of this vector,
+and expand its divergence using the following vector identity:
 
 $$\begin{gathered}
     \nabla \cdot (\vb{U} \: f)
     = \vb{U} \cdot (\nabla f) + (\nabla \cdot \vb{U}) \: f
-    \\
+\end{gathered}$$
+
+Inserting the scalar $f = h_2 h_3 V_1$
+the vector $\vb{U} = \vu{e}_1 / (h_2 h_3)$,
+we arrive at:
+
+$$\begin{gathered}
     \nabla \cdot \Big( \frac{\vu{e}_1}{h_2 h_3} (h_2 h_3 V_1) \Big)
     = \frac{\vu{e}_1}{h_2 h_3} \cdot \Big( \nabla (h_2 h_3 V_1) \Big)
     + \Big( \nabla \cdot \frac{\vu{e}_1}{h_2 h_3} \Big) (h_2 h_3 V_1)
 \end{gathered}$$
 
-The first term is straightforward to calculate
-thanks to our preceding expression for the gradient.
+The first right-hand term is easy to calculate
+thanks to our expression for the gradient $\nabla f$.
 Only the $\vu{e}_1$-component survives due to the dot product:
 
 $$\begin{aligned}
@@ -162,8 +191,8 @@ $$\begin{aligned}
     = \frac{\vu{e}_1}{h_1 h_2 h_3} \pdv{(h_2 h_3 V_1)}{x_1}
 \end{aligned}$$
 
-The second term is a bit more involved.
-To begin with, we use the gradient formula to note that:
+The second term is more involved.
+First, we use the gradient formula to observe that:
 
 $$\begin{aligned}
     \nabla x_1
@@ -177,7 +206,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Because $\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$ in an orthogonal basis,
-we can get the vector whose divergence we want:
+these gradients can be used to express the vector whose divergence we want:
 
 $$\begin{aligned}
     \nabla x_2 \cross \nabla x_3
@@ -196,15 +225,9 @@ $$\begin{aligned}
 \end{aligned}$$
 
 After repeating this procedure for the other components of $\vb{V}$,
-we arrive at the following general expression for the divergence $\nabla \cdot \vb{V}$:
-
-$$\begin{aligned}
-    \boxed{
-        \nabla \cdot \vb{V}
-        = \frac{1}{h_1 h_2 h_3}
-        \Big( \pdv{(h_2 h_3 V_1)}{x_1} + \pdv{(h_1 h_3 V_2)}{x_2} + \pdv{(h_1 h_2 V_3)}{x_3} \Big)
-    }
-\end{aligned}$$
+we get the desired general expression for the divergence.
+</div>
+</div>
 
 
 ## Laplacian
@@ -229,31 +252,55 @@ $$\begin{aligned}
 
 ## Curl
 
-We find the curl in a similar way as the divergence.
-Consider an arbitrary vector $\vb{V}$:
+The curl of a vector $\vb{V}$ is as follows
+in a general orthogonal curvilinear system:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \nabla \times \vb{V}
+            &= \frac{\vu{e}_1}{h_2 h_3} \Big( \pdv{(h_3 V_3)}{x_2} - \pdv{(h_2 V_2)}{x_3} \Big)
+            \\
+            &+ \frac{\vu{e}_2}{h_1 h_3} \Big( \pdv{(h_1 V_1)}{x_3} - \pdv{(h_3 V_3)}{x_1} \Big)
+            \\
+            &+ \frac{\vu{e}_3}{h_1 h_2} \Big( \pdv{(h_2 V_2)}{x_1} - \pdv{(h_1 V_1)}{x_2} \Big)
+        \end{aligned}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-curl"/>
+<label for="proof-curl">Proof</label>
+<div class="hidden">
+<label for="proof-curl">Proof.</label>
+The curl is found in a similar way as the divergence.
+We rewrite $\vb{V}$ like so:
 
 $$\begin{aligned}
     \vb{V}
-    = \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3
     = \frac{\vu{e}_1}{h_1} (h_1 V_1) + \frac{\vu{e}_2}{h_2} (h_2 V_2) + \frac{\vu{e}_3}{h_3} (h_3 V_3)
 \end{aligned}$$
 
-We expand the curl of its $\vu{e}_1$-component using a vector identity,
-where $f = h_1 V_1$ is a scalar and $\vb{U} = \vu{e}_1 / h_1$ is a vector:
+We expand the curl of its $\vu{e}_1$-component using the following vector identity:
 
 $$\begin{gathered}
     \nabla \cross (\vb{U} \: f)
     = (\nabla \cross \vb{U}) \: f - \vb{U} \cross (\nabla f)
-    \\
+\end{gathered}$$
+
+Inserting the scalar $f = h_1 V_1$
+and the vector $\vb{U} = \vu{e}_1 / h_1$, we arrive at:
+
+$$\begin{gathered}
     \nabla \cross \Big( \frac{\vu{e}_1}{h_1} (h_1 V_1) \Big)
     = \Big( \nabla \cross \frac{\vu{e}_1}{h_1} \Big) (h_1 V_1) - \frac{\vu{e}_1}{h_1} \cross \Big( \nabla (h_1 V_1) \Big)
 \end{gathered}$$
 
-Previously, when calculating the divergence,
+Previously, when proving the divergence,
 we already showed that $\vu{e}_1 / h_1 = \nabla x_1$.
 Because the curl of a gradient is zero,
-the first term thus disappears, leaving only the second,
-which contains a gradient turning out to be:
+the first term disappears, leaving only the second,
+which contains a gradient that turns out to be:
 
 $$\begin{aligned}
     \nabla (h_1 V_1)
@@ -273,20 +320,9 @@ $$\begin{aligned}
 \end{aligned}$$
 
 If we go through the same process for the other components of $\vb{V}$
-and add the results together, we get the following expression for the curl $\nabla \cross \vb{V}$:
-
-$$\begin{aligned}
-    \boxed{
-        \begin{aligned}
-            \nabla \times \vb{V}
-            &= \frac{\vu{e}_1}{h_2 h_3} \Big( \pdv{(h_3 V_3)}{x_2} - \pdv{(h_2 V_2)}{x_3} \Big)
-            \\
-            &+ \frac{\vu{e}_2}{h_1 h_3} \Big( \pdv{(h_1 V_1)}{x_3} - \pdv{(h_3 V_3)}{x_1} \Big)
-            \\
-            &+ \frac{\vu{e}_3}{h_1 h_2} \Big( \pdv{(h_2 V_2)}{x_1} - \pdv{(h_1 V_1)}{x_2} \Big)
-        \end{aligned}
-    }
-\end{aligned}$$
+and add up the results, we get the desired expression for the curl.
+</div>
+</div>
 
 
 ## Differential elements