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1 files changed, 52 insertions, 51 deletions

diff --git a/source/know/concept/curvilinear-coordinates/index.md b/source/know/concept/curvilinear-coordinates/index.md
index 3012ca6..cb22e43 100644
--- a/source/know/concept/curvilinear-coordinates/index.md
+++ b/source/know/concept/curvilinear-coordinates/index.md
@@ -14,15 +14,15 @@ is known as a **coordinate surface**, and the intersections of
 the surfaces of different coordinates are called **coordinate lines**.
 
 A **curvilinear** coordinate system is one where at least one of the coordinate surfaces is curved,
-e.g. in cylindrical coordinates the line between $r$ and $z$ is a circle.
+e.g. in cylindrical coordinates the line between $$r$$ and $$z$$ is a circle.
 If the coordinate surfaces are mutually perpendicular,
 it is an **orthogonal** system, which is generally desirable.
 
-A useful attribute of a coordinate system is its **line element** $\dd{\ell}$,
+A useful attribute of a coordinate system is its **line element** $$\dd{\ell}$$,
 which represents the differential element of a line in any direction.
-For an orthogonal system, its square $\dd{\ell}^2$ is calculated
-by taking the differential elements of the old Cartesian $(x, y, z)$ system
-and writing them out in the new $(x_1, x_2, x_3)$ system.
+For an orthogonal system, its square $$\dd{\ell}^2$$ is calculated
+by taking the differential elements of the old Cartesian $$(x, y, z)$$ system
+and writing them out in the new $$(x_1, x_2, x_3)$$ system.
 The resulting expression will be of the form:
 
 $$\begin{aligned}
@@ -33,7 +33,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where $h_1$, $h_2$, and $h_3$ are called **scale factors**,
+Where $$h_1$$, $$h_2$$, and $$h_3$$ are called **scale factors**,
 and need not be constants.
 The equation above only contains quadratic terms
 because the coordinate system is orthogonal by assumption.
@@ -45,20 +45,20 @@ and [cylindrical parabolic coordinates](/know/concept/cylindrical-parabolic-coor
 
 In the following subsections,
 we derive general formulae to convert expressions
-from Cartesian coordinates to 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
 
-Consider the the vector form of the line element $\dd{\ell}$,
-denoted by $\dd{\vu{\ell}}$ and expressed as:
+Consider the the vector form of the line element $$\dd{\ell}$$,
+denoted by $$\dd{\vu{\ell}}$$ and expressed as:
 
 $$\begin{aligned}
     \dd{\vu{\ell}}
     = \vu{e}_x \dd{x} + \vu{e}_y \dd{y} + \vu{e}_z \dd{z}
 \end{aligned}$$
 
-We can expand the Cartesian differential elements, e.g. $\dd{y}$,
+We can expand the Cartesian differential elements, e.g. $$\dd{y}$$,
 in the new basis as follows:
 
 $$\begin{aligned}
@@ -66,17 +66,17 @@ $$\begin{aligned}
     = \pdv{y}{x_1} \dd{x_1} + \pdv{y}{x_2} \dd{x_2} + \pdv{y}{x_3} \dd{x_3}
 \end{aligned}$$
 
-If we write this out for $\dd{x}$, $\dd{y}$ and $\dd{z}$,
-and group the terms according to $\dd{x}_1$, $\dd{x}_2$ and $\dd{x}_3$,
-we can compare it the alternative form of $\dd{\vu{\ell}}$:
+If we write this out for $$\dd{x}$$, $$\dd{y}$$ and $$\dd{z}$$,
+and group the terms according to $$\dd{x}_1$$, $$\dd{x}_2$$ and $$\dd{x}_3$$,
+we can compare it the alternative form of $$\dd{\vu{\ell}}$$:
 
 $$\begin{aligned}
     \dd{\vu{\ell}}
     = \vu{e}_1 \:h_1 \dd{x_1} + \vu{e}_2 \:h_2 \dd{x_2} + \vu{e}_3 \:h_3 \dd{x_4}
 \end{aligned}$$
 
-From this, we can read off $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$.
-Here we only give $\vu{e}_1$, since $\vu{e}_2$ and $\vu{e}_3$ are analogous:
+From this, we can read off $$\vu{e}_1$$, $$\vu{e}_2$$ and $$\vu{e}_3$$.
+Here we only give $$\vu{e}_1$$, since $$\vu{e}_2$$ and $$\vu{e}_3$$ are analogous:
 
 $$\begin{aligned}
     \boxed{
@@ -89,9 +89,9 @@ $$\begin{aligned}
 ## Gradient
 
 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 $x_3$:
+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 $$x_3$$:
 
 $$\begin{gathered}
     \boxed{
@@ -107,8 +107,8 @@ $$\begin{gathered}
 <label for="proof-grad">Proof</label>
 <div class="hidden" markdown="1">
 <label for="proof-grad">Proof.</label>
-For a direction $\dd{\ell}$, we know that
-$\idv{f}{\ell}$ is the component of $\nabla f$ in that direction:
+For a direction $$\dd{\ell}$$, we know that
+$$\idv{f}{\ell}$$ is the component of $$\nabla f$$ in that direction:
 
 $$\begin{aligned}
     \dv{f}{\ell}
@@ -117,9 +117,9 @@ $$\begin{aligned}
     = \nabla f \cdot \vu{u}
 \end{aligned}$$
 
-Where $\vu{u}$ is simply a unit vector in the direction of $\dd{\ell}$.
-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:
+Where $$\vu{u}$$ is simply a unit vector in the direction of $$\dd{\ell}$$.
+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:
 
 $$\begin{gathered}
     \nabla f
@@ -127,13 +127,14 @@ $$\begin{gathered}
     + \vu{e}_2 \dv{x_2}{\ell} \pdv{f}{x_2}
     + \vu{e}_3 \dv{x_3}{\ell} \pdv{f}{x_3}
 \end{gathered}$$
+
 </div>
 </div>
 
 
 ## Divergence
 
-The divergence of a vector $\vb{V} = \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3$
+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:
 
 $$\begin{aligned}
@@ -149,7 +150,7 @@ $$\begin{aligned}
 <label for="proof-div">Proof</label>
 <div class="hidden" markdown="1">
 <label for="proof-div">Proof.</label>
-As preparation, we rewrite $\vb{V}$ as follows
+As preparation, we rewrite $$\vb{V}$$ as follows
 to introduce the scale factors:
 
 $$\begin{aligned}
@@ -159,7 +160,7 @@ $$\begin{aligned}
     + \vu{e}_3 \frac{1}{h_1 h_2} (h_1 h_2 V_3)
 \end{aligned}$$
 
-We start by taking only the $\vu{e}_1$-component of this vector,
+We start by taking only the $$\vu{e}_1$$-component of this vector,
 and expand its divergence using the following vector identity:
 
 $$\begin{gathered}
@@ -167,8 +168,8 @@ $$\begin{gathered}
     = \vb{U} \cdot (\nabla f) + (\nabla \cdot \vb{U}) f
 \end{gathered}$$
 
-Inserting the scalar $f = h_2 h_3 V_1$
-the vector $\vb{U} = \vu{e}_1 / (h_2 h_3)$,
+Inserting the scalar $$f = h_2 h_3 V_1$$
+the vector $$\vb{U} = \vu{e}_1 / (h_2 h_3)$$,
 we arrive at:
 
 $$\begin{gathered}
@@ -178,8 +179,8 @@ $$\begin{gathered}
 \end{gathered}$$
 
 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:
+thanks to our expression for the gradient $$\nabla f$$.
+Only the $$\vu{e}_1$$-component survives due to the dot product:
 
 $$\begin{aligned}
     \frac{\vu{e}_1}{h_2 h_3} \cdot \Big( \nabla (h_2 h_3 V_1) \Big)
@@ -200,7 +201,7 @@ $$\begin{aligned}
     = \frac{\vu{e}_3}{h_3}
 \end{aligned}$$
 
-Because $\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$ in an orthogonal basis,
+Because $$\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$$ in an orthogonal basis,
 these gradients can be used to express the vector whose divergence we want:
 
 $$\begin{aligned}
@@ -210,7 +211,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 We then apply the divergence and expand the expression using a vector identity.
-In all cases, the curl of a gradient $\nabla \cross \nabla f$ is zero, so:
+In all cases, the curl of a gradient $$\nabla \cross \nabla f$$ is zero, so:
 
 $$\begin{aligned}
     \nabla \cdot \frac{\vu{e}_1}{h_2 h_3}
@@ -219,7 +220,7 @@ $$\begin{aligned}
     = 0
 \end{aligned}$$
 
-After repeating this procedure for the other components of $\vb{V}$,
+After repeating this procedure for the other components of $$\vb{V}$$,
 we get the desired general expression for the divergence.
 </div>
 </div>
@@ -227,7 +228,7 @@ we get the desired general expression for the divergence.
 
 ## Laplacian
 
-The Laplacian $\nabla^2 f$ is simply $\nabla \cdot \nabla f$,
+The Laplacian $$\nabla^2 f$$ is simply $$\nabla \cdot \nabla f$$,
 so we can find the general formula
 by combining the two preceding results
 for the gradient and the divergence:
@@ -247,7 +248,7 @@ $$\begin{aligned}
 
 ## Curl
 
-The curl of a vector $\vb{V}$ is as follows
+The curl of a vector $$\vb{V}$$ is as follows
 in a general orthogonal curvilinear system:
 
 $$\begin{aligned}
@@ -269,22 +270,22 @@ $$\begin{aligned}
 <div class="hidden" markdown="1">
 <label for="proof-curl">Proof.</label>
 The curl is found in a similar way as the divergence.
-We rewrite $\vb{V}$ like so:
+We rewrite $$\vb{V}$$ like so:
 
 $$\begin{aligned}
     \vb{V}
     = \frac{\vu{e}_1}{h_1} (h_1 V_1) + \frac{\vu{e}_2}{h_2} (h_2 V_2) + \frac{\vu{e}_3}{h_3} (h_3 V_3)
 \end{aligned}$$
 
-We expand the curl of its $\vu{e}_1$-component using the following vector identity:
+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:
+Inserting the scalar $$f = h_1 V_1$$
+and the vector $$\vb{U} = \vu{e}_1 / h_1$$, we arrive at:
 
 $$\begin{gathered}
     \nabla \cross \Big( \frac{\vu{e}_1}{h_1} (h_1 V_1) \Big)
@@ -292,7 +293,7 @@ $$\begin{gathered}
 \end{gathered}$$
 
 Previously, when proving the divergence,
-we already showed that $\vu{e}_1 / h_1 = \nabla x_1$.
+we already showed that $$\vu{e}_1 / h_1 = \nabla x_1$$.
 Because the curl of a gradient is zero,
 the first term disappears, leaving only the second,
 which contains a gradient that turns out to be:
@@ -304,8 +305,8 @@ $$\begin{aligned}
     + \vu{e}_3 \frac{1}{h_3} \pdv{(h_1 V_1)}{x_3}
 \end{aligned}$$
 
-Consequently, the curl of the first component of $\vb{V}$ is as follows,
-using the fact that $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$
+Consequently, the curl of the first component of $$\vb{V}$$ is as follows,
+using the fact that $$\vu{e}_1$$, $$\vu{e}_2$$ and $$\vu{e}_3$$
 are related to each other by cross products:
 
 $$\begin{aligned}
@@ -314,7 +315,7 @@ $$\begin{aligned}
     = - \frac{\vu{e}_3}{h_1 h_2} \pdv{(h_1 V_1)}{x_2} + \frac{\vu{e}_2}{h_1 h_3} \pdv{(h_1 V_1)}{x_3}
 \end{aligned}$$
 
-If we go through the same process for the other components of $\vb{V}$
+If we go through the same process for the other components of $$\vb{V}$$
 and add up the results, we get the desired expression for the curl.
 </div>
 </div>
@@ -322,14 +323,14 @@ and add up the results, we get the desired expression for the curl.
 
 ## Differential elements
 
-The point of the scale factors $h_1$, $h_2$ and $h_3$, as can seen from their derivation,
+The point of the scale factors $$h_1$$, $$h_2$$ and $$h_3$$, as can seen from their derivation,
 is to correct for "distortions" of the coordinates compared to the Cartesian system,
-such that the line element $\dd{\ell}$ retains its length.
-This property extends to the surface $\dd{S}$ and volume $\dd{V}$.
+such that the line element $$\dd{\ell}$$ retains its length.
+This property extends to the surface $$\dd{S}$$ and volume $$\dd{V}$$.
 
 When handling a differential volume in curvilinear coordinates,
 e.g. for a volume integral,
-the size of the box $\dd{V}$ must be corrected by the scale factors:
+the size of the box $$\dd{V}$$ must be corrected by the scale factors:
 
 $$\begin{aligned}
     \boxed{
@@ -339,8 +340,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-The same is true for the isosurfaces $\dd{S_1}$, $\dd{S_2}$ and $\dd{S_3}$
-where the coordinates $x_1$, $x_2$ and $x_3$ are respectively kept constant:
+The same is true for the isosurfaces $$\dd{S_1}$$, $$\dd{S_2}$$ and $$\dd{S_3}$$
+where the coordinates $$x_1$$, $$x_2$$ and $$x_3$$ are respectively kept constant:
 
 $$\begin{aligned}
     \boxed{
@@ -354,7 +355,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Using the same logic, the normal vector element $\dd{\vu{S}}$
+Using the same logic, the normal vector element $$\dd{\vu{S}}$$
 of an arbitrary surface is given by:
 
 $$\begin{aligned}
@@ -364,7 +365,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Finally, the tangent vector element $\dd{\vu{\ell}}$ takes the following form:
+Finally, the tangent vector element $$\dd{\vu{\ell}}$$ takes the following form:
 
 $$\begin{aligned}
     \boxed{