From 86ac658c6c23e5cad97010641a67ce00a9a76b40 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Mon, 29 May 2023 18:06:30 +0200 Subject: Rewrite and rename "Curvilinear coordinates" --- .../know/concept/curvilinear-coordinates/index.md | 375 --------------------- 1 file changed, 375 deletions(-) delete mode 100644 source/know/concept/curvilinear-coordinates/index.md (limited to 'source/know/concept/curvilinear-coordinates') 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. -- cgit v1.2.3