--- 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.