From 6ce0bb9a8f9fd7d169cbb414a9537d68c5290aae Mon Sep 17 00:00:00 2001 From: Prefetch Date: Fri, 14 Oct 2022 23:25:28 +0200 Subject: Initial commit after migration from Hugo --- .../concept/cylindrical-polar-coordinates/index.md | 200 +++++++++++++++++++++ 1 file changed, 200 insertions(+) create mode 100644 source/know/concept/cylindrical-polar-coordinates/index.md (limited to 'source/know/concept/cylindrical-polar-coordinates') diff --git a/source/know/concept/cylindrical-polar-coordinates/index.md b/source/know/concept/cylindrical-polar-coordinates/index.md new file mode 100644 index 0000000..a91e53e --- /dev/null +++ b/source/know/concept/cylindrical-polar-coordinates/index.md @@ -0,0 +1,200 @@ +--- +title: "Cylindrical polar coordinates" +date: 2021-07-26 +categories: +- Mathematics +- Physics +layout: "concept" +--- + +**Cylindrical polar coordinates** are an extension of polar coordinates to 3D, +which describes the location of a point in space +using the coordinates $(r, \varphi, z)$. +The $z$-axis is unchanged from Cartesian coordinates, +hence it is called a *cylindrical* system. + +Cartesian coordinates $(x, y, z)$ +and the cylindrical system $(r, \varphi, z)$ are related by: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + x &= r \cos\varphi \\ + y &= r \sin\varphi \\ + z &= z + \end{aligned} + } +\end{aligned}$$ + +Conversely, a point given in $(x, y, z)$ +can be converted to $(r, \varphi, z)$ +using these formulae: + +$$\begin{aligned} + \boxed{ + r = \sqrt{x^2 + y^2} + \qquad + \varphi = \mathtt{atan2}(y, x) + \qquad + z = z + } +\end{aligned}$$ + +The cylindrical polar coordinates form an orthogonal +[curvilinear system](/know/concept/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: + +$$\begin{aligned} + \dd{x} = \dd{r} \cos\varphi - \dd{\varphi} r \sin\varphi + \qquad + \dd{y} = \dd{r} \sin\varphi + \dd{\varphi} r \cos\varphi + \qquad + \dd{z} = \dd{z} +\end{aligned}$$ + +And then we calculate the line element $\dd{\ell}^2$, +skipping many terms thanks to orthogonality, + +$$\begin{aligned} + \dd{\ell}^2 + &= \dd{r}^2 \big( \cos^2(\varphi) + \sin^2(\varphi) \big) + + \dd{\varphi}^2 \big( r^2 \sin^2(\varphi) + r^2 \cos^2(\varphi) \big) + + \dd{z}^2 + \\ + &= \dd{r}^2 + r^2 \: \dd{\varphi}^2 + \dd{z}^2 +\end{aligned}$$ + +Finally, we can simply read off +the squares of the desired scale factors +$h_r^2$, $h_\varphi^2$ and $h_z^2$: + +$$\begin{aligned} + \boxed{ + h_r = 1 + \qquad + h_\varphi = r + \qquad + h_z = 1 + } +\end{aligned}$$ + +With these factors, we can easily convert things from the Cartesian system +using the standard formulae for orthogonal curvilinear coordinates. +The basis vectors are: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + \vu{e}_r + &= \cos\varphi \:\vu{e}_x + \sin\varphi \:\vu{e}_y + \\ + \vu{e}_\varphi + &= - \sin\varphi \:\vu{e}_x + \cos\varphi \:\vu{e}_y + \\ + \vu{e}_z + &= \vu{e}_z + \end{aligned} + } +\end{aligned}$$ + +The basic vector operations (gradient, divergence, Laplacian and curl) are given by: + +$$\begin{aligned} + \boxed{ + \nabla f + = \vu{e}_r \pdv{f}{r} + + \vu{e}_\varphi \frac{1}{r} \pdv{f}{\varphi} + + \mathbf{e}_z \pdv{f}{z} + } +\end{aligned}$$ + +$$\begin{aligned} + \boxed{ + \nabla \cdot \vb{V} + = \frac{1}{r} \pdv{(r V_r)}{r} + + \frac{1}{r} \pdv{V_\varphi}{\varphi} + + \pdv{V_z}{z} + } +\end{aligned}$$ + +$$\begin{aligned} + \boxed{ + \nabla^2 f + = \frac{1}{r} \pdv{}{r}\Big( r \pdv{f}{r} \Big) + + \frac{1}{r^2} \pdvn{2}{f}{\varphi} + + \pdvn{2}{f}{z} + } +\end{aligned}$$ + +$$\begin{aligned} + \boxed{ + \begin{aligned} + \nabla \times \vb{V} + &= \vu{e}_r \Big( \frac{1}{r} \pdv{V_z}{\varphi} - \pdv{V_\varphi}{z} \Big) + \\ + &+ \vu{e}_\varphi \Big( \pdv{V_r}{z} - \pdv{V_z}{r} \Big) + \\ + &+ \frac{\vu{e}_z}{r} \Big( \pdv{(r V_\varphi)}{r} - \pdv{V_r}{\varphi} \Big) + \end{aligned} + } +\end{aligned}$$ + +The differential element of volume $\dd{V}$ +takes the following form: + +$$\begin{aligned} + \boxed{ + \dd{V} + = r \dd{r} \dd{\varphi} \dd{z} + } +\end{aligned}$$ + +So, for example, an integral over all of space is converted like so: + +$$\begin{aligned} + \iiint_{-\infty}^\infty f(x, y, z) \dd{V} + = \int_{-\infty}^{\infty} \int_0^{2\pi} \int_0^\infty f(r, \varphi, z) \: r \dd{r} \dd{\varphi} \dd{z} +\end{aligned}$$ + +The isosurface elements are as follows, where $S_r$ is a surface at constant $r$, etc.: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + \dd{S}_r = r \dd{\varphi} \dd{z} + \qquad + \dd{S}_\varphi = \dd{r} \dd{z} + \qquad + \dd{S}_z = r \dd{r} \dd{\varphi} + \end{aligned} + } +\end{aligned}$$ + +Similarly, the normal vector element $\dd{\vu{S}}$ for an arbitrary surface is given by: + +$$\begin{aligned} + \boxed{ + \dd{\vu{S}} + = \vu{e}_r \: r \dd{\varphi} \dd{z} + + \vu{e}_\varphi \dd{r} \dd{z} + + \vu{e}_z \: r \dd{r} \dd{\varphi} + } +\end{aligned}$$ + +And finally, the tangent vector element $\dd{\vu{\ell}}$ of a given curve is as follows: + +$$\begin{aligned} + \boxed{ + \dd{\vu{\ell}} + = \vu{e}_r \dd{r} + + \vu{e}_\varphi \: r \dd{\varphi} + + \vu{e}_z \dd{z} + } +\end{aligned}$$ + + +## References +1. M.L. Boas, + *Mathematical methods in the physical sciences*, 2nd edition, + Wiley. -- cgit v1.2.3