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----
-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.