From e12c7ce372ecaa042d85d9fb76371a75ff518d1a Mon Sep 17 00:00:00 2001 From: Prefetch Date: Wed, 28 Jul 2021 14:27:37 +0200 Subject: Expand knowledge base, fix a:visited CSS --- .../cylindrical-parabolic-coordinates/index.pdc | 188 +++++++++++++++++++++ 1 file changed, 188 insertions(+) create mode 100644 content/know/concept/cylindrical-parabolic-coordinates/index.pdc (limited to 'content/know/concept/cylindrical-parabolic-coordinates') diff --git a/content/know/concept/cylindrical-parabolic-coordinates/index.pdc b/content/know/concept/cylindrical-parabolic-coordinates/index.pdc new file mode 100644 index 0000000..3460de4 --- /dev/null +++ b/content/know/concept/cylindrical-parabolic-coordinates/index.pdc @@ -0,0 +1,188 @@ +--- +title: "Cylindrical parabolic coordinates" +firstLetter: "C" +publishDate: 2021-03-04 +categories: +- Mathematics +- Physics + +date: 2021-03-04T15:07:46+01:00 +draft: false +markup: pandoc +--- + +# Cylindrical parabolic coordinates + +**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: + +$$\begin{aligned} + 2 x + = \frac{y^2}{\sigma^2} - \sigma^2 + \qquad \quad + 2 x + = - \frac{y^2}{\tau^2} + \tau^2 +\end{aligned}$$ + +Cylindrical parabolic coordinates form an orthogonal +[curvilinear](/know/concept/curvilinear-coordinates/) system, +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: + +$$\begin{aligned} + \boxed{ + \nabla f + = \frac{\mathbf{e}_\sigma}{\sqrt{\sigma^2 + \tau^2}} \pdv{f}{\sigma} + + \frac{\mathbf{e}_\tau}{\sqrt{\sigma^2 + \tau^2}} \pdv{f}{\tau} + + \mathbf{e}_z \pdv{f}{z} + } +\end{aligned}$$ + +$$\begin{aligned} + \boxed{ + \nabla \cdot \mathbf{V} + = \frac{1}{\sigma^2 + \tau^2} + \Big( \pdv{(V_\sigma \sqrt{\sigma^2 + \tau^2})}{d\sigma} + \pdv{(V_\tau \sqrt{\sigma^2 + \tau^2})}{d\tau} \Big) + \pdv{V_z}{z} + } +\end{aligned}$$ + +$$\begin{aligned} + \boxed{ + \nabla^2 f + = \frac{1}{\sigma^2 + \tau^2} \Big( \pdv[2]{f}{\sigma} + \pdv[2]{f}{\tau} \Big) + \pdv[2]{f}{z} + } +\end{aligned}$$ + +$$\begin{aligned} + \boxed{ + \begin{aligned} + \nabla \times \mathbf{V} + &= \mathbf{e}_\sigma \Big( \frac{\mathbf{e}_1}{\sqrt{\sigma^2 + \tau^2}} \pdv{V_z}{\tau} - \pdv{V_\tau}{z} \Big) + \\ + &+ \mathbf{e}_\tau \Big( \pdv{V_\sigma}{z} - \frac{1}{\sqrt{\sigma^2 + \tau^2}} \pdv{V_z}{\sigma} \Big) + \\ + &+ \frac{\mathbf{e}_z}{\sigma^2 + \tau^2} + \Big( \pdv{(V_\tau \sqrt{\sigma^2 + \tau^2})}{\sigma} - \pdv{(V_\sigma \sqrt{\sigma^2 + \tau^2})}{\tau} \Big) + \end{aligned} + } +\end{aligned}$$ + +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: + +$$\begin{aligned} + \boxed{ + \dd{\vu{S}} + = \mathbf{e}_\sigma \sqrt{\sigma^2 + \tau^2} \dd{\tau} \dd{z} + + \mathbf{e}_\tau \sqrt{\sigma^2 + \tau^2} \dd{\sigma} \dd{z} + + \mathbf{e}_z (\sigma^2 + \tau^2) \dd{\sigma} \dd{\tau} + } +\end{aligned}$$ + +$$\begin{aligned} + \boxed{ + \dd{\vu{\ell}} + = \mathbf{e}_\sigma \sqrt{\sigma^2 + \tau^2} \dd{\sigma} + + \mathbf{e}_\tau \sqrt{\sigma^2 + \tau^2} \dd{\tau} + + \mathbf{e}_z \dd{z} + } +\end{aligned}$$ + + +## References +1. M.L. Boas, + *Mathematical methods in the physical sciences*, 2nd edition, + Wiley. -- cgit v1.2.3