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 --- .../cylindrical-parabolic-coordinates/index.md | 182 +++++++++++++++++++++ 1 file changed, 182 insertions(+) create mode 100644 source/know/concept/cylindrical-parabolic-coordinates/index.md (limited to 'source/know/concept/cylindrical-parabolic-coordinates') diff --git a/source/know/concept/cylindrical-parabolic-coordinates/index.md b/source/know/concept/cylindrical-parabolic-coordinates/index.md new file mode 100644 index 0000000..f9d0475 --- /dev/null +++ b/source/know/concept/cylindrical-parabolic-coordinates/index.md @@ -0,0 +1,182 @@ +--- +title: "Cylindrical parabolic coordinates" +date: 2021-03-04 +categories: +- Mathematics +- Physics +layout: "concept" +--- + +**Cylindrical parabolic coordinates** are a coordinate system +that describes a point in space using three coordinates $(\sigma, \tau, z)$. +The $z$-axis is unchanged from the Cartesian system, +hence it is called a *cylindrical* system. +In the $z$-isoplane, however, confocal parabolas are used. +These coordinates can be converted to the Cartesian $(x, y, z)$ as follows: + +$$\begin{aligned} + \boxed{ + x = \frac{1}{2} (\tau^2 - \sigma^2 ) + \qquad + y = \sigma \tau + \qquad + z = z + } +\end{aligned}$$ + +Converting the other way is a bit trickier. +It can be done by solving the following equations, +and potentially involves some fiddling with signs: + +$$\begin{aligned} + 2 x + = \frac{y^2}{\sigma^2} - \sigma^2 + \qquad \qquad + 2 x + = - \frac{y^2}{\tau^2} + \tau^2 +\end{aligned}$$ + +Cylindrical parabolic coordinates form an orthogonal +[curvilinear system](/know/concept/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: + +$$\begin{aligned} + \dd{x} = - \sigma \dd{\sigma} + \tau \dd{\tau} + \qquad + \dd{y} = \tau \dd{\sigma} + \sigma \dd{\tau} + \qquad + \dd{z} = \dd{z} +\end{aligned}$$ + +We calculate the line segment $\dd{\ell}^2$, +skipping many terms thanks to orthogonality: + +$$\begin{aligned} + \dd{\ell}^2 + &= (\sigma^2 + \tau^2) \:\dd{\sigma}^2 + (\tau^2 + \sigma^2) \:\dd{\tau}^2 + \dd{z}^2 +\end{aligned}$$ + +From this, we can directly read off the scale factors $h_\sigma^2$, $h_\tau^2$ and $h_z^2$, +which turn out to be: + +$$\begin{aligned} + \boxed{ + h_\sigma = \sqrt{\sigma^2 + \tau^2} + \qquad + h_\tau = \sqrt{\sigma^2 + \tau^2} + \qquad + h_z = 1 + } +\end{aligned}$$ + +With these scale factors, we can use +the general formulae for orthogonal curvilinear coordinates +to easily to convert things from the Cartesian system. +The basis vectors are: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + \vu{e}_\sigma + &= \frac{- \sigma}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_x + \frac{\tau}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_y + \\ + \vu{e}_\tau + &= \frac{\tau}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_x + \frac{\sigma}{\sqrt{\sigma^2 + \tau^2}} \vu{e}_y + \\ + \vu{e}_z + &= \vu{e}_z + \end{aligned} + } +\end{aligned}$$ + +The basic vector operations (gradient, divergence, Laplacian and curl) are given by: + +$$\begin{aligned} + \boxed{ + \nabla f + = \frac{\vu{e}_\sigma}{\sqrt{\sigma^2 + \tau^2}} \pdv{f}{\sigma} + + \frac{\vu{e}_\tau}{\sqrt{\sigma^2 + \tau^2}} \pdv{f}{\tau} + + \vu{e}_z \pdv{f}{z} + } +\end{aligned}$$ + +$$\begin{aligned} + \boxed{ + \nabla \cdot \vb{V} + = \frac{1}{\sigma^2 + \tau^2} + \Big( \pdv{(V_\sigma \sqrt{\sigma^2 + \tau^2})}{\sigma} + \pdv{(V_\tau \sqrt{\sigma^2 + \tau^2})}{\tau} \Big) + \pdv{V_z}{z} + } +\end{aligned}$$ + +$$\begin{aligned} + \boxed{ + \nabla^2 f + = \frac{1}{\sigma^2 + \tau^2} \Big( \pdvn{2}{f}{\sigma} + \pdvn{2}{f}{\tau} \Big) + \pdvn{2}{f}{z} + } +\end{aligned}$$ + +$$\begin{aligned} + \boxed{ + \begin{aligned} + \nabla \times \vb{V} + &= \vu{e}_\sigma \Big( \frac{\vu{e}_1}{\sqrt{\sigma^2 + \tau^2}} \pdv{V_z}{\tau} - \pdv{V_\tau}{z} \Big) + \\ + &+ \vu{e}_\tau \Big( \pdv{V_\sigma}{z} - \frac{1}{\sqrt{\sigma^2 + \tau^2}} \pdv{V_z}{\sigma} \Big) + \\ + &+ \frac{\vu{e}_z}{\sigma^2 + \tau^2} + \Big( \pdv{(V_\tau \sqrt{\sigma^2 + \tau^2})}{\sigma} - \pdv{(V_\sigma \sqrt{\sigma^2 + \tau^2})}{\tau} \Big) + \end{aligned} + } +\end{aligned}$$ + +The differential element of volume $\dd{V}$ +in cylindrical parabolic coordinates is given by: + +$$\begin{aligned} + \boxed{ + \dd{V} = (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau} \dd{z} + } +\end{aligned}$$ + +The differential elements of the isosurfaces are as follows, +where $\dd{S_\sigma}$ is the $\sigma$-isosurface, etc.: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + \dd{S_\sigma} &= \sqrt{\sigma^2 + \tau^2} \dd{\tau} \dd{z} + \\ + \dd{S_\tau} &= \sqrt{\sigma^2 + \tau^2} \dd{\sigma} \dd{z} + \\ + \dd{S_z} &= (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau} + \end{aligned} + } +\end{aligned}$$ + +The normal element $\dd{\vu{S}}$ of a surface and +the tangent element $\dd{\vu{\ell}}$ of a curve are respectively: + +$$\begin{aligned} + \boxed{ + \dd{\vu{S}} + = \vu{e}_\sigma \sqrt{\sigma^2 + \tau^2} \dd{\tau} \dd{z} + + \vu{e}_\tau \sqrt{\sigma^2 + \tau^2} \dd{\sigma} \dd{z} + + \vu{e}_z (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau} + } +\end{aligned}$$ + +$$\begin{aligned} + \boxed{ + \dd{\vu{\ell}} + = \vu{e}_\sigma \sqrt{\sigma^2 + \tau^2} \dd{\sigma} + + \vu{e}_\tau \sqrt{\sigma^2 + \tau^2} \dd{\tau} + + \vu{e}_z \dd{z} + } +\end{aligned}$$ + + +## References +1. M.L. Boas, + *Mathematical methods in the physical sciences*, 2nd edition, + Wiley. -- cgit v1.2.3