Rewrite and rename "Curvilinear coordinates"

5 files changed, 1033 insertions, 381 deletions

diff --git a/source/know/concept/curvilinear-coordinates/index.md b/source/know/concept/curvilinear-coordinates/index.md
deleted file mode 100644
index 48a5a72..0000000
--- a/source/know/concept/curvilinear-coordinates/index.md
+++ /dev/null
@@ -1,375 +0,0 @@
----
-title: "Curvilinear coordinates"
-sort_title: "Curvilinear coordinates"
-date: 2021-03-03
-categories:
-- Mathematics
-- Physics
-layout: "concept"
----
-
-In a 3D coordinate system, the isosurface of a coordinate
-(i.e. the surface where that coordinate is constant while the others vary)
-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.
-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** $$\dd{\ell}$$,
-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.
-The resulting expression will be of the form:
-
-$$\begin{aligned}
-    \boxed{
-        \dd{\ell}^2
-        = \dd{x}^2 + \dd{y}^2 + \dd{z}^2
-        = h_1^2 \dd{x_1}^2 + h_2^2 \dd{x_2}^2 + h_3^2 \dd{x_3}^2
-    }
-\end{aligned}$$
-
-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.
-
-Examples of orthogonal curvilinear coordinate systems include
-[spherical coordinates](/know/concept/spherical-coordinates/),
-[cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/),
-and [cylindrical parabolic coordinates](/know/concept/cylindrical-parabolic-coordinates/).
-
-In the following subsections,
-we derive general formulae to convert expressions
-from Cartesian coordinates to the new orthogonal system $$(x_1, x_2, x_3)$$.
-
-
-
-## Basis vectors
-
-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. $$\dd{y}$$,
-in the new basis as follows:
-
-$$\begin{aligned}
-    \dd{y}
-    = \pdv{y}{x_1} \dd{x_1} + \pdv{y}{x_2} \dd{x_2} + \pdv{y}{x_3} \dd{x_3}
-\end{aligned}$$
-
-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 $$\dd{\vu{\ell}}$$:
-
-$$\begin{aligned}
-    \dd{\vu{\ell}}
-    = \vu{e}_1 \:h_1 \dd{x_1} + \vu{e}_2 \:h_2 \dd{x_2} + \vu{e}_3 \:h_3 \dd{x_4}
-\end{aligned}$$
-
-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:
-
-$$\begin{aligned}
-    \boxed{
-        h_1 \vu{e}_1
-        = \vu{e}_x \pdv{x}{x_1} + \vu{e}_y \pdv{y}{x_1} + \vu{e}_z \pdv{y}{x_1}
-    }
-\end{aligned}$$
-
-
-
-## 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 $$x_3$$:
-
-$$\begin{gathered}
-    \boxed{
-        \nabla f
-        = \vu{e}_1 \frac{1}{h_1} \pdv{f}{x_1}
-        + \vu{e}_2 \frac{1}{h_2} \pdv{f}{x_2}
-        + \vu{e}_3 \frac{1}{h_3} \pdv{f}{x_3}
-    }
-\end{gathered}$$
-
-
-{% include proof/start.html id="proof-grad" -%}
-For a direction $$\dd{\ell}$$, we know that
-$$\idv{f}{\ell}$$ is the component of $$\nabla f$$ in that direction:
-
-$$\begin{aligned}
-    \dv{f}{\ell}
-    = \pdv{f}{x} \dv{x}{\ell} + \pdv{f}{y} \dv{y}{\ell} + \pdv{f}{z} \dv{z}{\ell}
-    = \nabla f \cdot \bigg( \dv{x}{\ell}, \dv{y}{\ell}, \dv{z}{\ell} \bigg)
-    = \nabla f \cdot \vu{u}
-\end{aligned}$$
-
-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:
-
-$$\begin{gathered}
-    \nabla f
-    = \vu{e}_1 \dv{x_1}{\ell} \pdv{f}{x_1}
-    + \vu{e}_2 \dv{x_2}{\ell} \pdv{f}{x_2}
-    + \vu{e}_3 \dv{x_3}{\ell} \pdv{f}{x_3}
-\end{gathered}$$
-{% include proof/end.html id="proof-grad" %}
-
-
-
-## Divergence
-
-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:
-
-$$\begin{aligned}
-    \boxed{
-        \nabla \cdot \vb{V}
-        = \frac{1}{h_1 h_2 h_3}
-        \Big( \pdv{(h_2 h_3 V_1)}{x_1} + \pdv{(h_1 h_3 V_2)}{x_2} + \pdv{(h_1 h_2 V_3)}{x_3} \Big)
-    }
-\end{aligned}$$
-
-
-{% include proof/start.html id="proof-div" -%}
-As preparation, we rewrite $$\vb{V}$$ as follows
-to introduce the scale factors:
-
-$$\begin{aligned}
-    \vb{V}
-    &= \vu{e}_1 \frac{1}{h_2 h_3} (h_2 h_3 V_1)
-    + \vu{e}_2 \frac{1}{h_1 h_3} (h_1 h_3 V_2)
-    + \vu{e}_3 \frac{1}{h_1 h_2} (h_1 h_2 V_3)
-\end{aligned}$$
-
-We 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 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:
-
-$$\begin{aligned}
-    \frac{\vu{e}_1}{h_2 h_3} \cdot \Big( \nabla (h_2 h_3 V_1) \Big)
-    = \frac{\vu{e}_1}{h_1 h_2 h_3} \pdv{(h_2 h_3 V_1)}{x_1}
-\end{aligned}$$
-
-The second term is more involved.
-First, we use the gradient formula to observe that:
-
-$$\begin{aligned}
-    \nabla x_1
-    = \frac{\vu{e}_1}{h_1}
-    \qquad \quad
-    \nabla x_2
-    = \frac{\vu{e}_2}{h_2}
-    \qquad \quad
-    \nabla x_3
-    = \frac{\vu{e}_3}{h_3}
-\end{aligned}$$
-
-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:
-
-$$\begin{aligned}
-    \nabla x_2 \cross \nabla x_3
-    = \frac{\vu{e}_2}{h_2} \cross \frac{\vu{e}_3}{h_3}
-    = \frac{\vu{e}_1}{h_2 h_3}
-\end{aligned}$$
-
-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:
-
-$$\begin{aligned}
-    \nabla \cdot \frac{\vu{e}_1}{h_2 h_3}
-    = \nabla \cdot \big( \nabla x_2 \cross \nabla x_3 \big)
-    = \nabla x_3 \cdot (\nabla \cross \nabla x_2) - \nabla x_2 \cdot (\nabla \cross \nabla x_3)
-    = 0
-\end{aligned}$$
-
-After repeating this procedure for the other components of $$\vb{V}$$,
-we get the desired general expression for the divergence.
-{% include proof/end.html id="proof-div" %}
-
-
-
-## Laplacian
-
-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:
-
-$$\begin{aligned}
-    \boxed{
-        \nabla^2 f
-        = \frac{1}{h_1 h_2 h_3}
-        \bigg(
-            \pdv{}{x_1}\Big(\! \frac{h_2 h_3}{h_1} \pdv{f}{x_1} \!\Big)
-            + \pdv{}{x_2}\Big(\! \frac{h_1 h_3}{h_2} \pdv{f}{x_2} \!\Big)
-            + \pdv{}{x_3}\Big(\! \frac{h_1 h_2}{h_3} \pdv{f}{x_3} \!\Big)
-        \bigg)
-    }
-\end{aligned}$$
-
-
-
-## Curl
-
-The curl of a vector $$\vb{V}$$ is as follows
-in a general orthogonal curvilinear system:
-
-$$\begin{aligned}
-    \boxed{
-        \begin{aligned}
-            \nabla \times \vb{V}
-            &= \frac{\vu{e}_1}{h_2 h_3} \Big( \pdv{(h_3 V_3)}{x_2} - \pdv{(h_2 V_2)}{x_3} \Big)
-            \\
-            &+ \frac{\vu{e}_2}{h_1 h_3} \Big( \pdv{(h_1 V_1)}{x_3} - \pdv{(h_3 V_3)}{x_1} \Big)
-            \\
-            &+ \frac{\vu{e}_3}{h_1 h_2} \Big( \pdv{(h_2 V_2)}{x_1} - \pdv{(h_1 V_1)}{x_2} \Big)
-        \end{aligned}
-    }
-\end{aligned}$$
-
-
-{% include proof/start.html id="proof-curl" -%}
-The curl is found in a similar way as the divergence.
-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:
-
-$$\begin{gathered}
-    \nabla \cross (\vb{U} \: f)
-    = (\nabla \cross \vb{U}) f - \vb{U} \cross (\nabla f)
-\end{gathered}$$
-
-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 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 disappears, leaving only the second,
-which contains a gradient that turns out to be:
-
-$$\begin{aligned}
-    \nabla (h_1 V_1)
-    = \vu{e}_1 \frac{1}{h_1} \pdv{(h_1 V_1)}{x_1}
-    + \vu{e}_2 \frac{1}{h_2} \pdv{(h_1 V_1)}{x_2}
-    + \vu{e}_3 \frac{1}{h_3} \pdv{(h_1 V_1)}{x_3}
-\end{aligned}$$
-
-Consequently, the curl of the first component of $$\vb{V}$$ is as follows,
-using the fact that $$\vu{e}_1$$, $$\vu{e}_2$$ and $$\vu{e}_3$$
-are related to each other by cross products:
-
-$$\begin{aligned}
-    \nabla \cross \Big( \frac{\vu{e}_1}{h_1} (h_1 V_1) \Big)
-    = - \frac{\vu{e}_1}{h_1} \cross \Big( \nabla (h_1 V_1) \Big)
-    = - \frac{\vu{e}_3}{h_1 h_2} \pdv{(h_1 V_1)}{x_2} + \frac{\vu{e}_2}{h_1 h_3} \pdv{(h_1 V_1)}{x_3}
-\end{aligned}$$
-
-If we go through the same process for the other components of $$\vb{V}$$
-and add up the results, we get the desired expression for the curl.
-{% include proof/end.html id="proof-curl" %}
-
-
-
-## Differential elements
-
-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 $$\dd{V}$$.
-
-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:
-
-$$\begin{aligned}
-    \boxed{
-        \dd{V}
-        = \dd{x}\dd{y}\dd{z}
-        = h_1 h_2 h_3 \dd{x_1} \dd{x_2} \dd{x_3}
-    }
-\end{aligned}$$
-
-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:
-
-$$\begin{aligned}
-    \boxed{
-        \begin{aligned}
-            \dd{S_1} &= h_2 h_3 \dd{x_2} \dd{x_3}
-            \\
-            \dd{S_2} &= h_1 h_3 \dd{x_1} \dd{x_3}
-            \\
-            \dd{S_3} &= h_1 h_2 \dd{x_1} \dd{x_2}
-        \end{aligned}
-    }
-\end{aligned}$$
-
-Using the same logic, the normal vector element $$\dd{\vu{S}}$$
-of an arbitrary surface is given by:
-
-$$\begin{aligned}
-    \boxed{
-        \dd{\vu{S}}
-        = \vu{e}_1 h_2 h_3 \dd{x_2} \dd{x_3} + \vu{e}_2 h_1 h_3 \dd{x_1} \dd{x_3} + \vu{e}_3 h_1 h_2 \dd{x_1} \dd{x_2}
-    }
-\end{aligned}$$
-
-Finally, the tangent vector element $$\dd{\vu{\ell}}$$ takes the following form:
-
-$$\begin{aligned}
-    \boxed{
-        \dd{\vu{\ell}}
-        = \vu{e}_1 h_1 \dd{x_1} + \vu{e}_2 h_2 \dd{x_2} + \vu{e}_3 h_3 \dd{x_3}
-    }
-\end{aligned}$$
-
-
-
-## References
-1.  M.L. Boas,
-    *Mathematical methods in the physical sciences*, 2nd edition,
-    Wiley.
diff --git a/source/know/concept/cylindrical-parabolic-coordinates/index.md b/source/know/concept/cylindrical-parabolic-coordinates/index.md
index c8e16da..766c9b6 100644
--- a/source/know/concept/cylindrical-parabolic-coordinates/index.md
+++ b/source/know/concept/cylindrical-parabolic-coordinates/index.md
@@ -37,8 +37,8 @@ $$\begin{aligned}
     = - \frac{y^2}{\tau^2} + \tau^2
 \end{aligned}$$
 
-Cylindrical parabolic coordinates form an orthogonal
-[curvilinear system](/know/concept/curvilinear-coordinates/),
+Cylindrical parabolic coordinates form
+an [orthogonal curvilinear system](/know/concept/orthogonal-curvilinear-coordinates/),
 so we would like to find its scale factors $$h_\sigma$$, $$h_\tau$$ and $$h_z$$.
 The differentials of the Cartesian coordinates are as follows:
 
diff --git a/source/know/concept/cylindrical-polar-coordinates/index.md b/source/know/concept/cylindrical-polar-coordinates/index.md
index 686a4ed..43b4684 100644
--- a/source/know/concept/cylindrical-polar-coordinates/index.md
+++ b/source/know/concept/cylindrical-polar-coordinates/index.md
@@ -41,8 +41,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-The cylindrical polar coordinates form an orthogonal
-[curvilinear system](/know/concept/curvilinear-coordinates/),
+The cylindrical polar coordinates form
+an [orthogonal curvilinear system](/know/concept/orthogonal-curvilinear-coordinates/),
 whose scale factors $$h_r$$, $$h_\varphi$$ and $$h_z$$ we want to find.
 To do so, we calculate the differentials of the Cartesian coordinates:
 
diff --git a/source/know/concept/orthogonal-curvilinear-coordinates/index.md b/source/know/concept/orthogonal-curvilinear-coordinates/index.md
new file mode 100644
index 0000000..4fb45b4
--- /dev/null
+++ b/source/know/concept/orthogonal-curvilinear-coordinates/index.md
@@ -0,0 +1,1027 @@
+---
+title: "Orthogonal curvilinear coordinates"
+sort_title: "Orthogonal curvilinear coordinates"
+date: 2023-05-29 # Originally 2021-03-03, major rewrite
+categories:
+- Mathematics
+- Physics
+layout: "concept"
+---
+
+In a 3D coordinate system, the isosurface of a coordinate
+(i.e. the surface where that coordinate is constant while the others vary)
+is known as a **coordinate surface**, and the intersection line
+of two coordinates surfaces is called a **coordinate line**.
+
+A **curvilinear** coordinate system is one
+where at least one of the coordinate surfaces is curved:
+e.g. in cylindrical coordinates, the coordinate line of $$r$$ and $$z$$ is a circle.
+Here we limit ourselves to **orthogonal** systems,
+where the coordinate surfaces are always perpendicular.
+Examples of such orthogonal curvilinear systems include
+[spherical coordinates](/know/concept/spherical-coordinates/),
+[cylindrical polar coordinates](/know/concept/cylindrical-polar-coordinates/),
+and [cylindrical parabolic coordinates](/know/concept/cylindrical-parabolic-coordinates/).
+
+
+
+## Scale factors and basis vectors
+
+Given such a system with coordinates $$(c_1, c_2, c_3)$$.
+Their definition lets us convert all positions
+to classic Cartesian coordinates $$(x, y, z)$$
+using functions $$x$$, $$y$$ and $$z$$:
+
+$$\begin{aligned}
+    x
+    &= x(c_1, c_2, c_3)
+    \\
+    y
+    &= y(c_1, c_2, c_3)
+    \\
+    z
+    &= z(c_1, c_2, c_3)
+\end{aligned}$$
+
+A useful attribute of a coordinate system is its **line element** $$\dd{\vu{\ell}}$$,
+which represents the differential element of a line in any direction.
+Let $$\vu{e}_x$$, $$\vu{e}_y$$ and $$\vu{e}_z$$ be the Cartesian basis unit vectors:
+
+$$\begin{aligned}
+    \dd{\vu{\ell}}
+    \equiv \vu{e}_x \dd{x} + \: \vu{e}_y \dd{y} + \: \vu{e}_z \dd{z}
+\end{aligned}$$
+
+The Cartesian differential elements can be rewritten
+in $$(c_1, c_2, c_3)$$ with the chain rule:
+
+$$\begin{aligned}
+    \dd{\vu{\ell}}
+    = \quad &\vu{e}_x \bigg( \pdv{x}{c_1} \dd{c_1} + \: \pdv{x}{c_2} \dd{c_2} + \: \pdv{x}{c_3} \dd{c_3} \!\bigg)
+    \\
+    + \: &\vu{e}_y \bigg( \pdv{y}{c_1} \dd{c_1} + \: \pdv{y}{c_2} \dd{c_2} + \: \pdv{y}{c_3} \dd{c_3} \!\bigg)
+    \\
+    + \: &\vu{e}_z \bigg( \pdv{z}{c_1} \dd{c_1} + \: \pdv{z}{c_2} \dd{c_2} + \: \pdv{z}{c_3} \dd{c_3} \!\bigg)
+    \\
+    = \quad &\bigg( \pdv{x}{c_1} \vu{e}_x + \pdv{y}{c_1} \vu{e}_y + \pdv{z}{c_1} \vu{e}_z \bigg) \dd{c_1}
+    \\
+    + &\bigg( \pdv{x}{c_2} \vu{e}_x + \pdv{y}{c_2} \vu{e}_y + \pdv{z}{c_2} \vu{e}_z \bigg) \dd{c_2}
+    \\
+    + &\bigg( \pdv{x}{c_3} \vu{e}_x + \pdv{y}{c_3} \vu{e}_y + \pdv{z}{c_3} \vu{e}_z \bigg) \dd{c_3}
+\end{aligned}$$
+
+From this we define the **scale factors** $$h_1$$, $$h_2$$ and $$h_3$$
+and **local basis vectors** $$\vu{e}_1$$, $$\vu{e}_2$$ and $$\vu{e}_3$$:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            h_1 \vu{e}_1
+            &\equiv \pdv{x}{c_1} \vu{e}_x + \pdv{y}{c_1} \vu{e}_y + \pdv{z}{c_1} \vu{e}_z
+            \\
+            h_2 \vu{e}_2
+            &\equiv \pdv{x}{c_2} \vu{e}_x + \pdv{y}{c_2} \vu{e}_y + \pdv{z}{c_2} \vu{e}_z
+            \\
+            h_3 \vu{e}_3
+            &\equiv \pdv{x}{c_3} \vu{e}_x + \pdv{y}{c_3} \vu{e}_y + \pdv{z}{c_3} \vu{e}_z
+        \end{aligned}
+    }
+\end{aligned}$$
+
+Where $$\vu{e}_1$$, $$\vu{e}_2$$ and $$\vu{e}_3$$ are normalized,
+and orthogonal for any orthogonal curvilinear system.
+They are called *local* basis vectors
+because they generally depend on $$(c_1, c_2, c_3)$$,
+i.e. their directions vary from position to position.
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \vu{e}_x
+            &\equiv \pdv{c_1}{x} h_1 \vu{e}_1 + \pdv{c_2}{x} h_2 \vu{e}_2 + \pdv{c_3}{x} h_3 \vu{e}_3
+            \\
+            \vu{e}_y
+            &\equiv \pdv{c_1}{y} h_1 \vu{e}_1 + \pdv{c_2}{y} h_2 \vu{e}_2 + \pdv{c_3}{y} h_3 \vu{e}_3
+            \\
+            \vu{e}_z
+            &\equiv \pdv{c_1}{z} h_1 \vu{e}_1 + \pdv{c_2}{z} h_2 \vu{e}_2 + \pdv{c_3}{z} h_3 \vu{e}_3
+        \end{aligned}
+    }
+\end{aligned}$$
+
+
+In the following subsections, we use the scale factors $$h_1$$, $$h_2$$ and $$h_3$$
+to derive general formulae for converting vector calculus
+from Cartesian coordinates to $$(c_1, c_2, c_3)$$.
+
+
+
+## Differential elements
+
+The point of the scale factors $$h_1$$, $$h_2$$ and $$h_3$$,
+as can be seen from their derivation,
+is to correct for "distortions" of the coordinates compared to the Cartesian system,
+such that the line element $$\dd{\vu{\ell}}$$ retains its length.
+As was already established above:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vu{\ell}}
+        = \vu{e}_1 h_1 \dd{c_1} + \: \vu{e}_2 h_2 \dd{c_2} + \: \vu{e}_3 h_3 \dd{c_3}
+    }
+\end{aligned}$$
+
+These terms are the differentials along each of the local basis vectors.
+Let us now introduce the following notation, e.g. for $$c_1$$:
+
+$$\begin{aligned}
+    \dd{}_1\!\vb{x}
+    \equiv \pdv{\vb{x}}{c_1} \dd{c_1}
+    = \Big( \pdv{x}{c_1} \vu{e}_x + \pdv{y}{c_1} \vu{e}_y + \pdv{z}{c_1} \vu{e}_z \Big) \dd{c_1}
+    = \vu{e}_1 h_1 \dd{c_1}
+\end{aligned}$$
+
+And likewise we define $$\dd{}_2\!\vb{x}$$ and $$\dd{}_3\!\vb{x}$$.
+All differential elements (as found in e.g. integrals)
+can be expressed in terms of $$\dd{}_1\!\vb{x}$$, $$\dd{}_2\!\vb{x}$$ and $$\dd{}_3\!\vb{x}$$.
+
+The differential normal vector element $$\dd{\vu{S}}$$ in a surface integral is hence given by:
+
+$$\begin{aligned}
+    \dd{\vu{S}}
+    &= \dd{}_1\!\vb{x} \cross \dd{}_2\!\vb{x} + \dd{}_2\!\vb{x} \cross \dd{}_3\!\vb{x} + \dd{}_3\!\vb{x} \cross \dd{}_1\!\vb{x}
+    \\
+    &= (\vu{e}_1 \cross \vu{e}_2) \: h_1 h_2 \dd{c_1} \dd{c_2}
+    + \: (\vu{e}_2 \cross \vu{e}_3) \: h_2 h_3 \dd{c_2} \dd{c_3}
+    + \: (\vu{e}_3 \cross \vu{e}_1) \: h_1 h_3 \dd{c_1} \dd{c_3}
+\end{aligned}$$
+
+In an orthonormal basis we have
+$$\vu{e}_1 \cross \vu{e}_2 = \vu{e}_3$$,
+$$\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$$ and
+$$\vu{e}_3 \cross \vu{e}_1 = \vu{e}_2$$, so:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{\vu{S}}
+        = \vu{e}_1 \: h_2 h_3 \dd{c_2} \dd{c_3} + \: \vu{e}_2 \: h_1 h_3 \dd{c_1} \dd{c_3} + \: \vu{e}_3 \: h_1 h_2 \dd{c_1} \dd{c_2}
+    }
+\end{aligned}$$
+
+Next, the differential volume $$\dd{V}$$
+must also be corrected by the scale factors:
+
+$$\begin{aligned}
+    \dd{V}
+    = \dd{}_1\!\vb{x} \cross \dd{}_2\!\vb{x} \cdot \dd{}_3\!\vb{x}
+    = (\vu{e}_1 \cross \vu{e}_2 \cdot \vu{e}_3) \: h_1 h_2 h_3 \dd{c_1} \dd{c_2} \dd{c_3}
+\end{aligned}$$
+
+Once again $$\vu{e}_1 \cross \vu{e}_2 = \vu{e}_3$$,
+so the vectors disappear from the expression, leaving:
+
+$$\begin{aligned}
+    \boxed{
+        \dd{V}
+        = h_1 h_2 h_3 \dd{c_1} \dd{c_2} \dd{c_3}
+    }
+\end{aligned}$$
+
+
+
+## Basis vector derivatives
+
+Orthonormality tells us that $$\vu{e}_j \cdot \vu{e}_j = 1$$ for $$j = 1,2,3$$.
+Differentiating with respect to $$c_k$$:
+
+$$\begin{aligned}
+    \pdv{}{c_k} (\vu{e}_j \cdot \vu{e}_j)
+    = 2 \pdv{\vu{e}_j}{c_k} \cdot \vu{e}_j
+    = \pdv{}{c_k} 1
+    = 0
+\end{aligned}$$
+
+This means that the $$c_k$$-derivative of $$\vu{e}_j$$
+will always be orthogonal to $$\vu{e}_j$$, for all $$j$$ and $$k$$.
+Indeed, the general expression for the derivative of a local basis vector is:
+
+$$\begin{aligned}
+    \boxed{
+        \pdv{\vu{e}_j}{c_k}
+        = \frac{1}{h_j} \pdv{h_k}{c_j} \vu{e}_k - \delta_{jk} \sum_{l} \frac{1}{h_l} \pdv{h_j}{c_l} \vu{e}_l
+    }
+\end{aligned}$$
+
+Where $$\delta_{jk}$$ is the Kronecker delta.
+For example, if $$j = 1$$, writing this out gives:
+
+$$\begin{aligned}
+    \pdv{\vu{e}_1}{c_1}
+    &= - \frac{1}{h_2} \pdv{h_1}{c_2} \vu{e}_2 - \frac{1}{h_3} \pdv{h_1}{c_3} \vu{e}_3
+    \\
+    \pdv{\vu{e}_1}{c_2}
+    &= \frac{1}{h_1} \pdv{h_2}{c_1} \vu{e}_2
+    \\
+    \pdv{\vu{e}_1}{c_3}
+    &= \frac{1}{h_1} \pdv{h_3}{c_1} \vu{e}_3
+\end{aligned}$$
+
+
+{% include proof/start.html id="proof-deriv-basis" -%}
+In this proof we set $$j = 1$$ and $$k = 2$$ for clarity,
+but the approach is valid for any $$j \neq k$$.
+We know the definitions of $$h_1 \vu{e}_1$$ and $$h_2 \vu{e}_2$$,
+and that differentiations can be reordered:
+
+$$\begin{aligned}
+    \pdv{}{c_2} (h_1 \vu{e}_1)
+    &= \pdv{}{c_2} \pdv{}{c_1} \big( x \vu{e}_x + y \vu{e}_y + z \vu{e}_z \big)
+    = \pdv{}{c_1} (h_2 \vu{e}_2)
+\end{aligned}$$
+
+Expanding this according to the product rule of differentiation:
+
+$$\begin{aligned}
+    \pdv{h_1}{c_2} \vu{e}_1 + h_1 \pdv{\vu{e}_1}{c_2}
+    = \pdv{h_2}{c_1} \vu{e}_2 + h_2 \pdv{\vu{e}_2}{c_1}
+\end{aligned}$$
+
+We rearrange this in two different ways.
+Indeed, these two equations are identical:
+
+$$\begin{aligned}
+    h_1 \pdv{\vu{e}_1}{c_2}
+    &= \pdv{h_2}{c_1} \vu{e}_2 + \Big( h_2 \pdv{\vu{e}_2}{c_1} - \pdv{h_1}{c_2} \vu{e}_1 \Big)
+    \\
+    h_2 \pdv{\vu{e}_2}{c_1}
+    &= \pdv{h_1}{c_2} \vu{e}_1 + \Big( h_1 \pdv{\vu{e}_1}{c_2} - \pdv{h_2}{c_1} \vu{e}_2 \Big)
+\end{aligned}$$
+
+Recall that all derivatives of $$\vu{e}_j$$ are orthogonal to $$\vu{e}_j$$.
+Therefore, the first equation's right-hand side must be orthogonal to $$\vu{e}_1$$,
+and the second's to $$\vu{e}_2$$.
+We deduce that the parenthesized expressions
+are proportional to $$\vu{e}_3$$,
+and call the proportionality factors $$\lambda_{123}$$ and $$\lambda_{213}$$:
+
+$$\begin{aligned}
+    h_1 \pdv{\vu{e}_1}{c_2}
+    &= \pdv{h_2}{c_1} \vu{e}_2 + \lambda_{213} \vu{e}_3
+    \\
+    h_2 \pdv{\vu{e}_2}{c_1}
+    &= \pdv{h_1}{c_2} \vu{e}_1 + \lambda_{123} \vu{e}_3
+\end{aligned}$$
+
+Since these equations are identical,
+by comparing the definition of $$\lambda_{123}$$ to the other side of the equation,
+we see that $$\lambda_{123} = \lambda_{213}$$:
+
+$$\begin{aligned}
+    \lambda_{123} \vu{e}_3
+    &= h_1 \pdv{\vu{e}_1}{c_2} - \pdv{h_2}{c_1} \vu{e}_2
+    = h_2 \pdv{\vu{e}_2}{c_1} - \pdv{h_1}{c_2} \vu{e}_1
+    = \lambda_{213} \vu{e}_3
+\end{aligned}$$
+
+In general, $$\lambda_{jkl} = \lambda_{kjl}$$ for $$j \neq k \neq l$$.
+Next, we dot-multiply $$\lambda_{123}$$'s equation by $$\vu{e}_3$$,
+using that $$\vu{e}_2 \cdot \vu{e}_3 = 0$$
+and consequently $$\ipdv{(\vu{e}_2 \cdot \vu{e}_3)}{c_1} = 0$$:
+
+$$\begin{aligned}
+    \lambda_{123}
+    &= h_2 \pdv{\vu{e}_2}{c_1} \cdot \vu{e}_3 - \pdv{h_1}{c_2} \vu{e}_1 \cdot \vu{e}_3
+    = h_2 \pdv{\vu{e}_2}{c_1} \cdot \vu{e}_3
+    = - h_2 \frac{h_3}{h_3} \pdv{\vu{e}_3}{c_1} \cdot \vu{e}_2
+    = - \frac{h_2}{h_3} \lambda_{132}
+\end{aligned}$$
+
+In general, $$\lambda_{jkl} = - h_k \lambda_{jlk} / h_l$$ for $$j \neq k \neq l$$.
+Combining this fact with $$\lambda_{jkl} = \lambda_{kjl}$$ gives:
+
+$$\begin{aligned}
+    \lambda_{jkl}
+    = - \frac{h_k}{h_l} \lambda_{jlk}
+    = - \frac{h_k}{h_l} \lambda_{ljk}
+    = \frac{h_k}{h_l} \frac{h_j}{h_k} \lambda_{lkj}
+    = \frac{h_j}{h_l} \lambda_{klj}
+    = - \frac{h_j}{h_l} \frac{h_l}{h_j} \lambda_{kjl}
+    = - \lambda_{jkl}
+\end{aligned}$$
+
+But $$\lambda_{jkl} = -\lambda_{jkl}$$ is only possible if $$\lambda_{jkl}$$ is zero.
+Thus $$\lambda_{123}$$'s equation reduces to:
+
+$$\begin{aligned}
+    h_2 \pdv{\vu{e}_2}{c_1}
+    &= \pdv{h_1}{c_2} \vu{e}_1
+    \qquad \implies \qquad
+    \pdv{\vu{e}_2}{c_1}
+    = \frac{1}{h_2} \pdv{h_1}{c_2} \vu{e}_1
+\end{aligned}$$
+
+This gives us the general expression for $$\ipdv{\vu{e}_j}{c_k}$$ when $$j \neq k$$,
+but what about $$j = k$$?
+Well, from orthogonality we know:
+
+$$\begin{aligned}
+    0
+    = \vu{e}_2 \cdot \vu{e}_1
+    = \pdv{}{c_1} (\vu{e}_2 \cdot \vu{e}_1)
+    = \pdv{\vu{e}_2}{c_1} \cdot \vu{e}_1 + \vu{e}_2 \cdot \pdv{\vu{e}_1}{c_1}
+\end{aligned}$$
+
+We just calculated one of those terms, so this equation gives us the other:
+
+$$\begin{aligned}
+    \vu{e}_2 \cdot \pdv{\vu{e}_1}{c_1}
+    = - \pdv{\vu{e}_2}{c_1} \cdot \vu{e}_1
+    = -  \frac{1}{h_2} \pdv{h_1}{c_2}
+\end{aligned}$$
+
+Now we have the $$\vu{e}_2$$-component of $$\ipdv{\vu{e}_1}{c_1}$$,
+and can find the $$\vu{e}_3$$-component in the same way:
+
+$$\begin{aligned}
+    \vu{e}_3 \cdot \pdv{\vu{e}_1}{c_1}
+    = - \pdv{\vu{e}_3}{c_1} \cdot \vu{e}_1
+    = -  \frac{1}{h_3} \pdv{h_1}{c_3}
+\end{aligned}$$
+
+Adding up the $$\vu{e}_2$$- and $$\vu{e}_3$$-components gives the desired formula.
+There is no $$\vu{e}_1$$-component
+because $$\ipdv{\vu{e}_1}{c_1}$$ must be orthogonal to $$\vu{e}_1$$.
+{% include proof/end.html id="proof-deriv-basis" -%}
+
+
+
+## Gradient of a scalar
+
+In $$(c_1, c_2, c_3)$$, the gradient $$\nabla f$$ of a scalar field $$f$$
+has the following components:
+
+$$\begin{aligned}
+    \boxed{
+        (\nabla f)_j
+        = \frac{1}{h_j} \pdv{f}{c_j}
+    }
+\end{aligned}$$
+
+When this index notation is written out in full,
+the gradient $$\nabla f$$ becomes:
+
+$$\begin{aligned}
+    \nabla f
+    = \frac{1}{h_1} \pdv{f}{c_1} \vu{e}_1
+    + \frac{1}{h_2} \pdv{f}{c_2} \vu{e}_2
+    + \frac{1}{h_3} \pdv{f}{c_3} \vu{e}_3
+\end{aligned}$$
+
+
+{% include proof/start.html id="proof-grad-scalar" -%}
+For any unit vector $$\vu{u}$$, we can project $$\nabla f$$ onto it
+to get the component of $$\nabla f$$ along $$\vu{u}$$.
+Let us choose $$\vu{u} = \vu{e}_1$$, then such a projection gives:
+
+$$\begin{aligned}
+    \nabla f \cdot \vu{e}_1
+    &= \bigg( \pdv{f}{x} \vu{e}_x + \pdv{f}{y} \vu{e}_y + \pdv{f}{z} \vu{e}_z \bigg)
+    \cdot \frac{1}{h_1} \bigg( \pdv{x}{c_1} \vu{e}_x + \pdv{y}{c_1} \vu{e}_y + \pdv{z}{c_1} \vu{e}_z \bigg)
+    \\
+    &= \frac{1}{h_1} \bigg( \pdv{f}{x} \pdv{x}{c_1} + \pdv{f}{y} \pdv{y}{c_1} + \pdv{f}{z} \pdv{z}{c_1} \bigg)
+    \\
+    &= \frac{1}{h_1} \pdv{f}{c_1}
+\end{aligned}$$
+
+And we can do the same for $$\vu{e}_2$$ and $$\vu{e}_3$$,
+yielding analogous results:
+
+$$\begin{aligned}
+    \nabla f \cdot \vu{e}_2
+    = \frac{1}{h_2} \pdv{f}{c_2}
+    \qquad \qquad
+    \nabla f \cdot \vu{e}_3
+    = \frac{1}{h_3} \pdv{f}{c_3}
+\end{aligned}$$
+
+Finally, to express $$\nabla f$$ in the new coordinate system $$(c_1, c_2, c_3)$$,
+we simply combine these projections for all the basis vectors:
+
+$$\begin{aligned}
+    \nabla f
+    = (\nabla f \cdot \vu{e}_1) \vu{e}_1
+    + (\nabla f \cdot \vu{e}_2) \vu{e}_2
+    + (\nabla f \cdot \vu{e}_3) \vu{e}_3
+\end{aligned}$$
+{% include proof/end.html id="proof-grad-scalar" %}
+
+
+
+## Divergence of a vector
+
+The divergence of a vector field $$\vb{V} = V_1 \vu{e}_1 + V_2 \vu{e}_2 + V_3 \vu{e}_3$$
+is given in $$(c_1, c_2, c_3)$$ by:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{V}
+        = \sum_{j} \frac{1}{H} \pdv{}{c_j} \bigg( \frac{H V_j}{h_j} \bigg)
+    }
+\end{aligned}$$
+
+Where $$H \equiv h_1 h_2 h_3$$.
+When this index notation is written out in full, it becomes:
+
+$$\begin{aligned}
+    \nabla \cdot \vb{V}
+    = \frac{1}{h_1 h_2 h_3}
+    \bigg( \pdv{}{c_1} (h_2 h_3 V_1) + \pdv{}{c_2} (h_1 h_3 V_2) + \pdv{}{c_3} (h_1 h_2 V_3) \bigg)
+\end{aligned}$$
+
+
+{% include proof/start.html id="proof-div-vector-1" label="Proof 1" -%}
+From our earlier calculation of $$\nabla f$$,
+we know how to express the del $$\nabla$$ in $$(c_1, c_2, c_3)$$.
+Now we simply take the dot product of $$\nabla$$ and $$\vb{V}$$:
+
+$$\begin{aligned}
+    \nabla \cdot \vb{V}
+    &= \bigg( \vu{e}_1 \frac{1}{h_1} \pdv{}{c_1} + \vu{e}_2 \frac{1}{h_2} \pdv{}{c_2} + \vu{e}_3 \frac{1}{h_3} \pdv{}{c_3} \bigg)
+    \cdot \bigg( V_1 \vu{e}_1 + V_2 \vu{e}_2 + V_3 \vu{e}_3 \bigg)
+    \\
+    &= \bigg( \sum_{j} \vu{e}_j \frac{1}{h_j} \pdv{}{c_j} \bigg) \cdot \bigg( \sum_{k} V_k \vu{e}_k \bigg)
+    \\
+    &= \sum_{jk} \vu{e}_j \cdot \frac{1}{h_j} \pdv{}{c_j} (V_k \vu{e}_k)
+    \\
+    &= \sum_{jk} (\vu{e}_j \cdot \vu{e}_k) \frac{1}{h_j} \pdv{V_k}{c_j}
+    + \sum_{jk} \Big( \vu{e}_j \cdot \pdv{\vu{e}_k}{c_j} \Big) \frac{V_k}{h_j}
+\end{aligned}$$
+
+Substituting our expression for the derivatives of the local basis vectors, we find:
+
+$$\begin{aligned}
+    \nabla \cdot \vb{V}
+    &= \sum_{jk} (\vu{e}_j \cdot \vu{e}_k) \frac{1}{h_j} \pdv{V_k}{c_j}
+    + \sum_{jk} \vu{e}_j
+    \cdot \bigg( \frac{1}{h_k} \pdv{h_j}{c_k} \vu{e}_j - \delta_{jk} \sum_{l} \frac{1}{h_l} \pdv{h_k}{c_l} \vu{e}_l \bigg) \frac{V_k}{h_j}
+    \\
+    &= \sum_{jk} (\vu{e}_j \cdot \vu{e}_k) \frac{1}{h_j} \pdv{V_k}{c_j}
+    + \sum_{jk} (\vu{e}_j \cdot \vu{e}_j) \frac{V_k}{h_j h_k} \pdv{h_j}{c_k}
+    - \sum_{jl} (\vu{e}_j \cdot \vu{e}_l) \frac{V_j}{h_j h_l} \pdv{h_j}{c_l}
+    \\
+    &= \sum_{j} \frac{1}{h_j} \pdv{V_j}{c_j}
+    + \sum_{jk} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k}
+    - \sum_{j} \frac{V_j}{h_j h_j} \pdv{h_j}{c_j}
+    \\
+    &= \sum_{j} \frac{1}{h_j} \pdv{V_j}{c_j}
+    + \sum_{j} \sum_{k \neq j} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k}
+    \\
+\end{aligned}$$
+
+Where we noticed that the latter two terms cancel out if $$k = j$$.
+Now, to proceed, it is easiest to just write out the index notation:
+
+$$\begin{aligned}
+    \nabla \cdot \vb{V}
+    &= \quad\: \sum_{j} \frac{1}{h_j} \pdv{V_j}{c_j} + \sum_{j} \sum_{k \neq j} \frac{V_k}{h_j h_k} \pdv{h_j}{c_k}
+    \\
+    &= \quad\: \frac{1}{h_1} \pdv{V_1}{c_1} + \frac{V_1}{h_1 h_2} \pdv{h_2}{c_1} + \frac{V_1}{h_1 h_3} \pdv{h_3}{c_1}
+    \\
+    &\quad\:\: + \frac{V_2}{h_1 h_2} \pdv{h_1}{c_2} + \frac{1}{h_2} \pdv{V_2}{c_2} + \frac{V_2}{h_2 h_3} \pdv{h_3}{c_2}
+    \\
+    &\quad\:\: + \frac{V_3}{h_1 h_3} \pdv{h_1}{c_3} + \frac{V_3}{h_2 h_3} \pdv{h_2}{c_3} + \frac{1}{h_3} \pdv{V_3}{c_3}
+    \\
+    &= \frac{1}{h_1 h_2 h_3} \bigg( h_2 h_3 \pdv{V_1}{c_1} + h_3 V_1 \pdv{h_2}{c_1} + h_2 V_1 \pdv{h_3}{c_1}
+    \\
+    &\qquad\qquad + h_3 V_2 \pdv{h_1}{c_2} + h_1 h_3 \pdv{V_2}{c_2} + h_1 V_2 \pdv{h_3}{c_2}
+    \\
+    &\qquad\qquad + h_2 V_3 \pdv{h_1}{c_3} + h_1 V_3 \pdv{h_2}{c_3} + h_1 h_2 \pdv{V_3}{c_3} \bigg)
+\end{aligned}$$
+
+Which can clearly be rewritten with the product rule,
+leading to the desired formula.
+{% include proof/end.html id="proof-div-vector-1" label="Proof 1" %}
+
+
+Boas gives an alternative proof, which is shorter but more specialized:
+
+
+{% include proof/start.html id="proof-div-vector-2" label="Proof 2" -%}
+We take the divergence of the $$c_1$$-component of $$\vb{V}$$ and expand it:
+
+$$\begin{aligned}
+    \nabla \cdot (V_1 \vu{e}_1)
+    &= \nabla \cdot \bigg( \Big( h_2 h_3 V_1 \Big) \Big( \frac{\vu{e}_1}{h_2 h_3} \Big) \bigg)
+    \\
+    &= \nabla (h_2 h_3 V_1) \cdot \Big( \frac{\vu{e}_1}{h_2 h_3} \Big)
+    + (h_2 h_3 V_1) \bigg( \nabla \cdot \Big( \frac{\vu{e}_1}{h_2 h_3} \Big) \bigg)
+\end{aligned}$$
+
+The latter term is zero, because
+in any orthogonal basis $$\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$$,
+and according to our gradient formula we have $$\nabla c_2 = \vu{e}_2 / h_2$$ etc., so:
+
+$$\begin{aligned}
+    \nabla \cdot \bigg( \frac{\vu{e}_1}{h_2 h_3} \bigg)
+    &= \nabla \cdot \bigg( \frac{\vu{e}_2}{h_2} \cross \frac{\vu{e}_3}{h_3} \bigg)
+    \\
+    &= \nabla \cdot \big( \nabla c_2 \cross \nabla c_3 \big)
+    \\
+    &= \nabla c_3 \cdot (\nabla \cross \nabla c_2) - \nabla c_2 \cdot (\nabla \cross \nabla c_3)
+    \\
+    &= 0
+\end{aligned}$$
+
+Where we used a vector identity and the fact that the curl of a gradient must vanish.
+We are thus left with the former term,
+to which we apply our gradient formula again,
+where only the $$\vu{e}_1$$-term survives due to the dot product and orthogonality:
+
+$$\begin{aligned}
+    \nabla \cdot (V_1 \vu{e}_1)
+    &= \nabla (h_2 h_3 V_1) \cdot \frac{\vu{e}_1}{h_2 h_3}
+    \\
+    &= \frac{1}{h_1 h_2 h_3} \pdv{}{c_1} (h_2 h_3 V_1)
+\end{aligned}$$
+
+We then repeat this procedure for $$\vb{V}$$'s other components,
+and simply add up the results to get the desired formula.
+{% include proof/end.html id="proof-div-vector-2" label="Proof 2" %}
+
+
+
+## Laplacian of a scalar
+
+The Laplacian $$\nabla^2 f$$ of a scala