From 9d741c2c762d8b629cef5ac5fbc26ca44c345a77 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Fri, 5 Mar 2021 16:41:32 +0100
Subject: Expand knowledge base

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+---
+title: "Curvilinear coordinates"
+firstLetter: "C"
+publishDate: 2021-03-03
+categories:
+- Mathematics
+- Physics
+
+date: 2021-03-03T19:47:34+01:00
+draft: false
+markup: pandoc
+---
+
+# Curvilinear coordinates
+
+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 coordinates,
+and [parabolic cylindrical coordinates](/know/concept/parabolic-cylindrical-coordinates/).
+
+In the following subsections,
+we derive general formulae to convert expressions
+from Cartesian coordinates in the new orthogonal system $(x_1, x_2, x_3)$.
+
+
+## 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
+
+For a given direction $\dd{\ell}$, we know that
+$\dv*{f}{\ell}$ 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 can thus find an 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:
+
+$$\begin{gathered}
+    \nabla f
+    = \vu{e}_1 \dv{x_1}{\ell} \pdv{f}{x_1}
+    + \vu{e}_2 \dv{x_2}{\ell} \pdv{f}{x_2}
+    + \vu{e}_3 \dv{x_3}{\ell} \pdv{f}{x_3}
+    \\
+    \boxed{
+        \nabla f
+        = \vu{e}_1 \frac{1}{h_1} \pdv{f}{x_1}
+        + \vu{e}_2 \frac{1}{h_2} \pdv{f}{x_2}
+        + \vu{e}_3 \frac{1}{h_3} \pdv{f}{x_3}
+    }
+\end{gathered}$$
+
+Where $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$
+are the basis unit vectors respectively corresponding to $x_1$, $x_2$ and $x_3$.
+
+
+## Divergence
+
+Consider a vector $\vb{V}$ in the target coordinate system
+with components $V_1$, $V_2$ and $V_3$:
+
+$$\begin{aligned}
+    \vb{V}
+    &= \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3
+    \\
+    &= \vu{e}_1 \frac{1}{h_2 h_3} (h_2 h_3 V_1)
+    + \vu{e}_2 \frac{1}{h_1 h_3} (h_1 h_3 V_2)
+    + \vu{e}_3 \frac{1}{h_1 h_2} (h_1 h_2 V_3)
+\end{aligned}$$
+
+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:
+
+$$\begin{gathered}
+    \nabla \cdot (\vb{U} \: f)
+    = \vb{U} \cdot (\nabla f) + (\nabla \cdot \vb{U}) \: f
+    \\
+    \nabla \cdot \Big( \frac{\vu{e}_1}{h_2 h_3} (h_2 h_3 V_1) \Big)
+    = \frac{\vu{e}_1}{h_2 h_3} \cdot \Big( \nabla (h_2 h_3 V_1) \Big)
+    + \Big( \nabla \cdot \frac{\vu{e}_1}{h_2 h_3} \Big) (h_2 h_3 V_1)
+\end{gathered}$$
+
+The first term is straightforward to calculate
+thanks to our preceding expression for the gradient.
+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 a bit more involved.
+To begin with, we use the gradient formula to note 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,
+we can get 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 arrive at the following general expression for the divergence $\nabla \cdot \vb{V}$:
+
+$$\begin{aligned}
+    \boxed{
+        \nabla \cdot \vb{V}
+        = \frac{1}{h_1 h_2 h_3}
+        \Big( \pdv{(h_2 h_3 V_1)}{x_1} + \pdv{(h_1 h_3 V_2)}{x_2} + \pdv{(h_1 h_2 V_3)}{x_3} \Big)
+    }
+\end{aligned}$$
+
+
+## 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
+
+We find the curl in a similar way as the divergence.
+Consider an arbitrary vector $\vb{V}$:
+
+$$\begin{aligned}
+    \vb{V}
+    = \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3
+    = \frac{\vu{e}_1}{h_1} (h_1 V_1) + \frac{\vu{e}_2}{h_2} (h_2 V_2) + \frac{\vu{e}_3}{h_3} (h_3 V_3)
+\end{aligned}$$
+
+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:
+
+$$\begin{gathered}
+    \nabla \cross (\vb{U} \: f)
+    = (\nabla \cross \vb{U}) \: f - \vb{U} \cross (\nabla f)
+    \\
+    \nabla \cross \Big( \frac{\vu{e}_1}{h_1} (h_1 V_1) \Big)
+    = \Big( \nabla \cross \frac{\vu{e}_1}{h_1} \Big) (h_1 V_1) - \frac{\vu{e}_1}{h_1} \cross \Big( \nabla (h_1 V_1) \Big)
+\end{gathered}$$
+
+Previously, when calculating the divergence,
+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:
+
+$$\begin{aligned}
+    \nabla (h_1 V_1)
+    = \vu{e}_1 \frac{1}{h_1} \pdv{(h_1 V_1)}{x_1}
+    + \vu{e}_2 \frac{1}{h_2} \pdv{(h_1 V_1)}{x_2}
+    + \vu{e}_3 \frac{1}{h_3} \pdv{(h_1 V_1)}{x_3}
+\end{aligned}$$
+
+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 the results together, we get the following expression for the curl $\nabla \cross \vb{V}$:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            \nabla \times \vb{V}
+            &= \frac{\vu{e}_1}{h_2 h_3} \Big( \pdv{(h_3 V_3)}{x_2} - \pdv{(h_2 V_2)}{x_3} \Big)
+            \\
+            &+ \frac{\vu{e}_2}{h_1 h_3} \Big( \pdv{(h_1 V_1)}{x_3} - \pdv{(h_3 V_3)}{x_1} \Big)
+            \\
+            &+ \frac{\vu{e}_3}{h_1 h_2} \Big( \pdv{(h_2 V_2)}{x_1} - \pdv{(h_1 V_1)}{x_2} \Big)
+        \end{aligned}
+    }
+\end{aligned}$$
+
+
+## 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.
-- 
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